Motion in One Dimension

Time

Reason — Time has only magnitude, whereas force, acceleration and displacement all have magnitude as well as direction.

Both (a) and (b)

Reason — Vector quantities have both magnitude and direction.

Velocity

Reason — A vector quantity has both magnitude and direction. Velocity has both magnitude and direction, so it is a vector quantity.

One dimensional

Reason — As the train is moving on a straight track and there is no lateral movement (sideways), hence it is a one dimensional motion.

0

Reason — If a body starts its motion from point A to B and comes back to the same point after a certain time interval, the displacement is 0. For example: In a circular motion, the body comes back to the starting point after completing the circle, then displacement is zero but distance covered is not zero.

Variable acceleration

Reason — As the change in velocity of the car in not same in same interval of time, the acceleration is said to be variable.

100 m

Reason — As the body is moving with uniform speed, equal distance is covered in equal intervals of time.

Distance covered in 1 s is 10 m, hence distance covered in 10 s is 10 x 10 = 100 m.

Displacement

Reason — If a body starts its motion from point A to B and comes back to the same after a certain time interval, the displacement is 0. Whereas, distance covered and speed will not be zero.

5 m s^-1

Reason — As,

$18 \text{ kmh}^{-1} = 18 \dfrac{\text{km}}{\text{h}} = \dfrac {18 \times 1000\ \text m}{60 \times 60\ \text s} \\[1em] \Rightarrow 18 \text{ kmh}^{-1} = \dfrac {18 \times 10 \text{m}}{6 \times 6 \text{s}} \\[1em] \Rightarrow 18 \text{\text{ kmh}}^{-1} = \dfrac {180 \text{ m}}{36 \text{ s}} \\[1em] \Rightarrow 18 \text{\text{ kmh}}^{-1} = 5 \text{ ms}^{-1}$

Hence, 18 km h^-1 is equal to 5 ms^-1.

All of these

Reason — The value of g does not depend on height, mass or shape of the body. On the earth's surface, g is maximum at the poles and minium at the equator. The value of g decreases with altitude and also with depth from the earth's surface.

Uniform acceleration

Reason — When a body is falling freely, the acceleration is said to be uniform as equal changes in velocity takes place in equal intervals of time.

Poles

Reason — On the earth's surface, g is maximum at the poles and minium at the equator. The value of g decreases with altitude and also with depth from the earth's surface.

The displacement is zero

Reason — The displacement is zero because the initial and the final position of the body is same.

The magnitude of velocity of a body in motion is its speed

Reason —

(1) Incorrect — Average speed is the total distance travelled divided by time, and is never zero unless the body doesn’t move at all. On the other hand, average velocity can be zero if the displacement is zero (like in a round trip).

(2) Incorrect — Speed is a scalar quantity, but velocity is a vector quantity as it has both magnitude and direction.

(3) Correct — The magnitude of velocity is the same as the speed of the body.

(4) Incorrect — Speed is always positive or zero (since it's a scalar quantity) but velocity can be positive or negative depending on the direction.

$\dfrac{5\text v_1 \text v_2}{3\text v_1 +2\text v_2}$

Reason — Let, total distance covered be S of which S₁ is covered with v₁ in time t₁ and S₂ with v₂ in time t₂.

Then,

$\text S_1=\dfrac{2}{5}\text S$ ............... (1)

and

$\text S_2=\dfrac{3}{5}\text S$ ............... (2)

As

$\text {Speed (v)}=\dfrac{\text {Distance (S)}}{\text {Time (t)}}$

Then,

$\Rightarrow \text v_1=\dfrac{\text S_1}{\text t_1} \\[1 em] \Rightarrow \text t_1=\dfrac{\text S_1}{\text v_1}$

Putting value of S₁ from equation 1

$\Rightarrow \text t_1 = \dfrac{2}{5}\dfrac{\text S}{\text v_1}$

Similarly

$\Rightarrow \text v_2=\dfrac{\text S_2}{\text t_2} \\ [1 em] \Rightarrow\text t_2=\dfrac{\text S_2}{\text v_2}$

Putting value of S₂ from equation 2

$\Rightarrow\text t_2= \dfrac{3}{5}\dfrac{\text S}{\text v_2}$ Now,

$\text {Average Speed (v)}=\dfrac{\text {Total distance travelled (S)}}{\text {Total time taken (t)}} \\[1 em] =\dfrac{\text S}{\text t_1 + \text t_2} \\[1 em] =\dfrac{\text S}{\dfrac{2\text S}{5\text v_1}+\dfrac{3\text S}{5\text v_2}} \\[1 em] = \dfrac{\text S}{\dfrac{\text S}{5}\Big(\dfrac{2}{\text v_1} + \dfrac{3}{\text v_2}\Big)} \\[1em] = \dfrac{\text S}{\dfrac{\text S}{5}\Big(\dfrac{2\text v_2 + 3 \text v_1}{\text v_1 \text v_2} \Big)} \\[1em] =\dfrac{5\text v_1 \text v_2}{3\text v_1 +2\text v_2}$

As,

$72 \text {km h}^{-1} = \dfrac{72 \text {km}}{1 \text {h}} = \dfrac {72 \times 1000 \text {m}}{(60 \times 60) \text {s}} \\[0.5em] \Rightarrow 72 \text {km h}^{-1} = \dfrac {72 \times 10 \text {m}}{(6 \times 6 )\text {s}} \\[0.5em] \Rightarrow 72 \text {\text {km h}}^{-1} = \dfrac {20 \times \text {m}}{\text {s}} \\[0.5em]$

Hence, 72km h^-1 is equal to 20 m s^-1.

As,

$15 \text {m s}^{-1} = 15 \dfrac{\text {m}}{\text {s}} = \dfrac {15 \times 60 \times 60 }{1000} \\[0.5em] \Rightarrow 15 \text {m s}^{-1} = \dfrac {15 \times 6 \times 6 }{10} \\[0.5em] \Rightarrow 15 \text {m s}^{-1}= 54 \text {km h}^{-1} \\[0.5em]$

Hence, 15 m s^-1 is equal to 54 kmh^-1.

(a) As,

$10 \text{ km h}^{-1} = \dfrac{10 \text{ km}}{1\text{h}} = \dfrac {10 \times 1000 \text{m}}{60 \times 60 \text{s}} \\[1em] \Rightarrow 10 \text{ km h}^{-1} = \dfrac {10 \times 10 \text{m}}{6 \times 6 \text{s}} \\[1em] \Rightarrow 10 \text{ km h}^{-1} = 2.78 \space \dfrac{\text{m}}{\text{s}}$

Hence, 10 km h^-1 is equal to 2.78 ms^-1.

(b) As,

$18 \text{ km min}^{-1} = \dfrac{18 \text{ km}}{1 \text{ min}} = \dfrac {18 \times 1000 \text{m}}{60 \text{s}} \\[1em] \Rightarrow 18 \text{ kmh}^{-1} = \dfrac {18 \times 100 \text{ m}}{6 \text{s}} \\[1em] \Rightarrow 18 \text{ kmh}^{-1}= 300 \space \dfrac{\text{ m}}{\text{s}}$

Hence, 18 km h^-1 is equal to 300 m s^-1.

In order to arrange in increasing order of speeds, the units of the given speeds must be same.

Hence, we convert all the three in m s^-1.

(a) 10 m s^-1 = 10 m s^-1   .... [1]
(already in m s^-1)

(b) 1 km min^-1,

$\text{1 km min}^{-1} = \dfrac{\text{1 km}}{\text{1 min}} = \dfrac {1 \times 1000 \text{m}}{ 60 \text{s}} \\[0.5em] \Rightarrow \text{1 km min}^{-1} = \dfrac {1 \times 100 \text{m}}{6\text{s}} \\[0.5em] \Rightarrow \text{1 km min}^{-1} = \dfrac {16.66 \times \text{m}}{\text{s}} \\[0.5em]$

Hence,
1 km min^-1 = 16.66 m s^-1   .... [2]

(c) 18 km h^-1,

As,

$\text{18 km h}^{-1} = 18 \dfrac{\text km}{\text h} = \dfrac {18 \times 1000 \text m}{60 \times 60 \text s} \\[0.5em] \Rightarrow \text{18 km h}^{-1} = \dfrac {18 \times 10 \text m}{6 \times 6\text s} \\[0.5em] \Rightarrow \text{18 km h}^{-1} = \dfrac {5 \times \text m}{\text s} \\[0.5em]$

Hence,
18km h^-1 = 5 m s^-1   .... [3]

From 1, 2 and 3, we get the increasing order as follows —

18 km h^-1, 10 m s^-1, 1 km min^-1.

As we know,

Distance = Speed x Time

Given,

time = 3 h

speed = 65 km h^-1

Substituting the values in the formula above, we get,

Distance = 65 x 3 = 195 km

Hence, the distance between the two cities = 195 km.

Given,

Distance travelled S₁ = 30 km

Speed v₁ = 60 km h^-1

Distance travelled S₂ = 30 km

Speed v₂ = 40 km h^-1

(i) As we know,

$\text {Time (t)} = \dfrac {\text {distance (S)}}{\text {speed (v)}}$

Substituting the values in the formula above, we get,

$\text{t}_1 = \dfrac{\text{S}_1}{\text{v}_1} \\[0.5em] \text{t}_1 = \dfrac{30 \text{km}}{60 \text{km h}^{-1}} \\[0.5em] \Rightarrow \text{t}_1 = \dfrac{1}{2} \text{h} \\[0.5em] \Rightarrow \text{t}_1 = 0.5 \text{h} \\[0.5em]$

Converting 0.5 h to min we get,

$\Rightarrow \text{t}_1 = 0.5 \times 60 \text{ min} \\[0.5em] \Rightarrow \text{t}_1 = 30 \text{ min} \\[0.5em]$

and

$\text{t}_2 = \dfrac{\text{S}_2}{\text{v}_2} \\[0.5em] \text{t}_2 = \dfrac{30 \text{km}}{40 \text{km h}^{-1}} \\[0.5em] \Rightarrow \text{t}_2 = \dfrac{3}{4} \text{h} \\[0.5em] \Rightarrow \text{t}_2 = 0.75 \text{h} \\[0.5em]$

Converting 0.75 h to min we get,

$\Rightarrow \text{t}_2 = 0.75 \times 60 \text{ min} \\[0.5em] \Rightarrow \text{t}_2 = 45 \text{ min} \\[0.5em]$

Total time t = t₁ + t₂
= 30 min + 45 min
= 75 min

Hence, Total time of journey = 75 min.

(ii) As we know,

Total distance S = S₁ + S₂
= 30 km + 30 km
= 60 km

Hence, Total distance travelled = 60 km.

As we know,

$\text{Average speed} = \dfrac{\text{total distance}}{\text{total time taken}} \\[0.5em] \text{Average speed} = \dfrac{60 \text { km}}{1.25 \text { h}} \\[0.5em] \Rightarrow \text{Average speed} = 48 \text { kmh} ^{-1}$

Hence, Average speed of the car = 48 kmh^-1

Given,

Distance travelled S₁ = 200 km

time t₁ = 2 h

Distance travelled S₂ = 200 km

time t₂ = 3 h

(i) As we know,

$\text{Average speed} = \dfrac{\text{total distance}}{\text{total time taken}} \\[0.5em]$

Total distance travelled S = S₁ + S₂
= 200 km + 200 km
= 400 km

Total time taken t = t₁ + t₂
= 2 h + 3 h
= 5 h

Substituting the values of total distance travelled and total time taken in the formula of Average Speed above, we get,

$\text{Average speed} = \dfrac{400 \text { km}}{5 \text { h}} \\[0.5em] \Rightarrow \text{Average speed} = 80 \text { kmh} ^{-1} \\[0.5em]$

Hence, Average speed of the car = 80 km h^-1

(ii) Displacement = 0 (since final position is same as initial position)

Hence, Average velocity = 0 ( as displacement = 0)

As we know,

$\text{Speed (v)} = \dfrac{\text{distance (S)}}{\text{time (t)}} \\[0.5em]$

Given,

Distance travelled S = 1 km = 1000 m,

time t = 100 s

Substituting the values in the formula above, we get,

$\text{Speed (v)} = \dfrac{1000 \text { m} }{ 100 \text{ s}} \\[0.5em] \Rightarrow \text{Speed (v)} = 10 \text { ms}^{-1} \\[0.5em]$

Hence, speed of the car is 10 m s^-1

(ii) The magnitude of velocity of the car is same as that of speed of car i.e., 10 m s^-1

However, when we talk about velocity, we mention direction also.

Hence, velocity of car is 10 m s^-1 due east.

Given,

Initial velocity (u) = 0

Final velocity (v) = 10 m s ^-1

Time (t) = 2 s

As we know,

$\text {a} = \dfrac {\text {final vel. (v)} - \text {initial vel. (u)}}{\text {time (t)}}$

Substituting the values in the formula above, we get,

$\text {Acceleration (a)} = \dfrac {10 - 0}{2} \\[0.5em] \text {Acceleration (a)} = \dfrac {10\text{ ms}^{-1}}{2 \text{ s}} \\[0.5em] \text {Acceleration (a)} = 5 \text {ms}^{-2} \\[0.5em]$

Hence, acceleration of the body is 5 m s ^-2.

As we know,

$\text {a} = \dfrac {\text {final vel. (v)} - \text {initial vel. (u)}}{\text {time (t)}}$

Given,

Initial velocity (u) = 0

Final velocity (v) = 180 m s ^-1

Time (t) = 0.05 h

Converting 0.05 h to s we get,

$0.05 \text{h} = 0.05 \times 60 \times 60 \\[0.5em] 0.05 \text{h} = 0.05 \times 3600 \\[0.5em] 0.05 \text{h} = 180 \text {s} \\[0.5em]$

Substituting the values in the formula above, we get,

$\text {Acceleration (a)} = \dfrac {180 - 0}{180} \\[0.5em] \text {Acceleration (a)} = \dfrac {180 \text { ms}^{-1}}{180 \text { s}} \\[0.5em] \Rightarrow \text {Acceleration (a)} = 1 \text { ms}^{-2} \\[0.5em]$

Hence, acceleration of the body is 1 m s ^-2.

As we know,

$\text {a} = \dfrac {\text {final vel. (v)} - \text {initial vel. (u)}}{\text {time (t)}}$

Given,

Initial velocity (u) = 50 m s ^-1

Final velocity (v) = 20 m s ^-1

Time (t) = 3 s

Substituting the values in the formula above, we get,

$\text {Acceleration (a)} = \dfrac {20 - 50}{3} \\[0.5em] \text {Acceleration (a)} = - \dfrac {30\text { ms}^{-1}}{3\text { s}} \\[0.5em] \Rightarrow \text {Acceleration (a)} = -10 \text { ms}^{-2} \\[0.5em]$

Hence, acceleration of the body is -10 m s ^-2. Negative sign shows that the velocity decreases with time, so retardation is 10 m s ^-2.

As we know,

$\text {a} = \dfrac {\text {final vel. (v)} - \text {initial vel. (u)}}{\text {time (t)}}$

Given,

Initial velocity (u) = 18 km h^-1

To convert initial velocity u to m s^-1

$18 \text { km h}^{-1} = \dfrac{18 \text { km}}{1 \text{ h}} = \dfrac {18 \times 1000 \text{ m}}{60 \times 60 \text{ s}} \\[0.5em] \Rightarrow 18 \text { km h}^{-1} = \dfrac {18 \times 10 \text{ m}}{6 \times 6 \text{ s}} \\[0.5em] \Rightarrow 18 \text { km h}^{-1} = 5 \text{ m s}^{-1} \\[0.5em]$

Hence, 18 km h^-1 is equal to 5 m s^-1.

Final velocity (v) = 0

Time (t) = 2 s

Substituting the values in the formula above, we get,

$\text {Acceleration (a)} = \dfrac {0 - 5}{2} \\[0.5em] \text {Acceleration (a)} = -\dfrac {5\text { ms}^{-1}}{2\text { s}} \\[0.5em] \Rightarrow \text {Acceleration (a)} = - 2.5 \text { ms}^{-2} \\[0.5em]$

Hence, acceleration of the body is -2.5 m s ^-2. Negative sign shows that the velocity decreases with time, so retardation is 2.5 m s ^-2.

As we know,

$\text {a} = \dfrac {\text {final vel. (v)} - \text {initial vel. (u)}}{\text {time (t)}}$

Hence,

${\text {Velocity Increase}} = \text {Acceleration (a)} \times {\text{time(t)}}$

Given,

a = 5 m s^-2

t = 2 s

Substituting the values in the formula above, we get,

$\text {Velocity Increase} = 5 \text{ m s}^{-2} \times 2 \text{ s} \\[0.5em] \Rightarrow \text {Velocity Increase} = 10 \text{ m s}^{-1} \\[0.5em]$

Hence, increase in velocity = 10 m s ^-1.

As we know,

$\text {a} = \dfrac {\text {final vel. (v)} - \text {initial vel. (u)}}{\text {time (t)}}$

and retardation is negative acceleration.

Hence,

$\text {final vel. (v)} - \text {initial vel. (u)} \\[0.5em] = \text {Retardation(-a)} \times {\text {time(t)}}$

Given,

a = - 2 m s^-2

t = 5 s

u = 20 m s ^-1

Substituting the values in the formula above, we get,

$\text {final vel. (v)} - 20 \text{m s}^{-1} = - 2 \text{m s}^{-2} \times 5 \text s \\[0.5em] \text {v} - 20 \text{m s}^{-1} = - 10 \text{m s}^{-1} \\[0.5em] \Rightarrow \text {v} = (20 - 10) \text{m s}^{-1} \\[0.5em] \Rightarrow \text {v} = 10 \text{m s}^{-1} \\[0.5em]$

Hence, final velocity = 10 m s ^-1.

As we know,

$\text {a} = \dfrac {\text {final vel. (v)} - \text {initial vel. (u)}}{\text {time (t)}}$

Hence,

$\text {final vel. (v)} - \text {initial vel. (u)} \\[0.5em] = \text {Acceleration (a)} \times {\text {time (t)}}$

Given,

a = 2 m s^-2

t = 5 s

u = 5.0 m s ^-1

Substituting the values in the formula above, we get,

$\text {final vel. (v)} - 5 \text{m s}^{-1} = 2 \text{m s}^{-2} \times 5 \text s \\[0.5em] \text {v} - 5 \text{m s}^{-1} = 10 \text{m s}^{-1} \\[0.5em] \Rightarrow \text {v} = (10 + 5) \text{m s}^{-1} \\[0.5em] \Rightarrow \text {v} = 15 \text{m s}^{-1} \\[0.5em]$

Hence, final velocity = 15 m s ^-1.

Given,

speed of car = 18 km h^-1

(i) To convert speed to m s^-1

$18 \text {km h}^{-1} = \dfrac{18 \text {km}}{1 \text{h}} = \dfrac {18 \times 1000 \text{m}}{60 \times 60 \text {s}} \\[0.5em] \Rightarrow 18 \text {km h}^{-1} = \dfrac {18 \times 10 \text {m}}{6 \times 6 \text {s}} \\[0.5em] \Rightarrow 18 \text {km h}^{-1} = 5 \text {m s}^{-1} \\[0.5em]$

Hence, 18 km h^-1 is equal to 5 m s^-1.

(ii) As the car is stopped, the final velocity (v) = 0

initial velocity (u) = 18 km h^-1

As we know,

$\text {a} = \dfrac {\text {final vel. (v)} - \text {initial vel. (u)}}{\text {time (t)}}$

Given,

Initial velocity (u) = 5 m s^-1

Final velocity (v) = 0

Time (t) = 5 s

Substituting the values in the formula above, we get,

$\text {Acceleration (a)} = \dfrac {0 - 5}{5} \\[0.5em] \text {Acceleration (a)} = -\dfrac {5}{5} \\[0.5em] \Rightarrow \text {Acceleration (a)} = - 1 \text {ms}^{-2} \\[0.5em]$

Hence, acceleration of the car is -1 m s^-2. Negative sign shows that the velocity decreases with time, so retardation is 1 m s ^-2.

(iii) Given,

t = 2 s

a = -1 m s^-2

u = 5 m s^-1

Substituting the values in the formula for acceleration, we get,

$- 1 \text {ms}^{-2} = \dfrac {(v - 5 ) \text {ms}^{-1}}{2 \text {s}} \\[0.5em] - 2 \text {ms}^{-1} = (v - 5) {\text {ms}}^{-1} \\[0.5em] \Rightarrow v = (5 - 2) \text {ms}^{-1} \\[0.5em] \Rightarrow v = 3 \text {ms}^{-1} \\[0.5em]$

Hence, the speed of car after 2 s of applying the brakes = 3 m s^-1.

Give an example of motion of a body moving with a constant speed, but with a variable velocity. Draw a diagram to represent such a motion.

The motion of a body in circular path, is an example of a body moving with a constant speed and variable velocity because the direction of motion of body changes continuously with time.

At any instant, the velocity is along the tangent to the circular path at that point.

The figure above shows the direction of velocity v at different points A, B, C and D of the circular path.

Differences between scalar and vector quantities are as follows —

No, the statement is not correct. Scalar quantities can be added, subtracted, multiplied and divided by simple arithmetic methods however vector quantities follow different algebra for their addition, subtraction and multiplication as they have magnitude as well as direction.

(a) The difference between distance and displacement is as follows :

(b) The difference between speed and velocity is as follows :

(c) The difference between uniform velocity and variable velocity is as follows :

(d) The difference between average speed and average velocity is as follows :

(e) The differences between acceleration and retardation are as follows :

(f) The differences between uniform acceleration and variable acceleration are as follows :

Yes, the displacement can be zero, even if the distance is not zero.

If a body, after travelling, comes back to its starting point, the displacement is zero but the distance travelled is not zero.

Example — When a body is thrown vertically upwards from a point A on the ground, after some time it comes back to the same point A, then the displacement of the body is zero but the distance travelled by the body is not zero.

The motion of a body in circular path, is an example of a body moving with a constant speed and variable velocity because the direction of motion of body changes continuously with time.

At any instant, the velocity is along the tangent to the circular path at that point.

The figure above shows the direction of velocity v at different points A, B, C and D of the circular path.

If a body starts its motion and comes back to the same point after a certain time, (e.g., in a circular motion,) the displacement is zero, so the average velocity is also zero, but the total distance travelled is not zero and therefore, the average speed is not zero.

(a) Uniform velocity — The rain drops reach on earth’s surface falling with uniform velocity.

(b) Variable velocity — The motion of a freely falling body is with variable velocity because although the direction of motion of the body does not change, but the speed continuously increases.

(c) Variable acceleration — The motion of a vehicle on a hilly (or crowded) road.

(d) Uniform retardation — Motion of a vehicle, reaching a destination.

With the help of the given diagram, we observe that the red dots (dripping oil) are initially at regular intervals and later the gap between the dots decreases, which implies that initially the car was covering equal distances in equal intervals of time but later the distance is decreasing.

Thus, we can infer that initially the car is moving with a constant speed and then it slows down.

The direction of velocity of an object moving in a circular path is always tangent to the circle. This means that with the movement of body the velocity changes.

No, the value of 'g' is not same at all places on the earth's surface. The value of g varies from place to place so an average value is considered. The average value of 'g' is 9.8 ms^-2 or nearly 10 ms^-2. The value of g decreases with altitude and also with depth from the earth's surface.

The value of 'g' is maximum at the poles and minimum at the equator on the earth surface.

Both will reach the ground simultaneously as value of g does not depend on the mass of the body. Acceleration due to gravity is same ( = g ) on both and there is no effect of friction and buoyancy due to air.

(a) Pressure — Scalar quantity

(b) Force — Vector quantity

(c) Momentum — Vector quantity

(d) Energy — Scalar quantity

(e) Weight — Vector quantity

(f) Speed — Scalar quantity

Two parameters are required to express a scalar quantity :

1. Unit in which the quantity is being expressed.
2. Numerical value of the measured quantity.

Three parameters are required to express a vector quantity :

1. Unit in which the quantity is being expressed.
2. Numerical value of the measured quantity.
3. Direction.

The negative sign with a vector quantity implies the reverse (or opposite) direction. e.g., the forces $\overrightarrow{F}$and $-\overrightarrow{F}$are in opposite directions.

A body is said to be at rest if it does not change its position with respect to its immediate surrounding.

A body is said to be in motion if it changes its position with respect to its immediate surroundings.

When a body moves along a straight line path, its motion is said to be one dimensional motion.

The shortest distance from the initial to the final position of the body, is the magnitude of displacement and its direction is from initial position to the final position. It is a vector quantity.

Unit — The S.I. unit of displacement is metre (m) and C.G.S. unit is centimetre (cm)

The magnitude of displacement is equal to the distance, when the motion is in a fixed direction.

The velocity of a body is the distance travelled per second by the body in a specified direction.

It is a vector quantity.

$\text{S.I. unit of velocity} = \dfrac{\text{metre}}{\text{second}} = \dfrac{\text{m}}{\text{s}} \\[0.5em] \text{C.G.S. unit of velocity} = \dfrac{\text{centimetre}}{\text{second}} = \dfrac{\text{cm}}{\text{s}} \\[0.5em]$.

Speed of a body is the rate of change of distance with time .

It is a scalar quantity. It is generally represented by u or v .

If a body travels a distance S in time t, then its speed v is —

$\text{Speed (v)} = \dfrac{\text{Distance (S)}}{\text{Time (t)}} = \dfrac{\text{S}}{\text{t}} \\[0.5em]$

$\text{S.I. unit of speed} = \dfrac{\text{metre}}{\text{second}} = \dfrac{\text{m}}{\text{s}} \\[0.5em] \text{C.G.S. unit of speed} = \dfrac{\text{centimetre}}{\text{second}} = \dfrac{\text{cm}}{\text{s}} \\[0.5em]$.

Velocity gives direction of motion of body as it is a vector quantity. Speed is a scalar quantity hence, it does not say anything about direction of motion.

In case of a body moving with uniform speed, the instantaneous speed and the average speed are equal (same as the uniform speed) because, in case of uniform speed, neither the speed nor the direction change.

Acceleration is the rate of change of velocity with time.

Numerically, acceleration is equal to the change in velocity in 1 s.

i.e.,

$\text{Acceleration} = \dfrac{\text{Change in velocity}}{\text{Time interval}} \\[0.5em] \text{S.I. unit of acceleration} = \dfrac{\text{meter}}{\text{second}^2} = \dfrac{\text{m}}{\text{s}^2}\\[0.5em] \text{C.G.S. unit of acceleration} = \dfrac{\text{centimeter}}{\text{second}^2} = \dfrac{\text{cm}}{\text{s}^2}\\[0.5em]$

If the velocity of a body decreases with time, it is called retardation. As it is decrease in velocity per second, so, retardation is negative acceleration.

It's unit is same as that of acceleration.

$\text{S.I. unit of retardation} = \dfrac{\text{meter}}{\text{second}^2} = \dfrac{\text{m}}{\text{s}^2}\\[0.5em] \text{C.G.S. unit of retardation} = \dfrac{\text{centimeter}}{\text{second}^2} = \dfrac{\text{cm}}{\text{s}^2}\\[0.5em]$

Velocity determines the direction of motion as velocity is the distance travelled per second by a moving object in a particular direction.

Positive or negative sign of velocity indicates the direction of motion.

Acceleration is the rate of change of velocity with time. It does not say anything about direction of motion.

Positive or negative sign of acceleration tell us whether the velocity is increasing or decreasing.

When a body falls freely under the gravity, the acceleration produced in the body due to earth's gravitational attraction is called acceleration due to gravity.

The average value of 'g' is 9.8 ms^-2 (or nearly 10 ms^-2).

(a) Below is the velocity-time graph for a body moving with uniform velocity :

Example : In the fig. given below, a straight line AB, represents the velocity-time graph of a body moving with a uniform velocity 4 ms^-1 for 5 s.

Displacement in 5 second = area of rectangle OABC = OC x OA = 5 x 4 = 20 m

Hence, displacement = 20m

The slope of the straight line AB is zero, therefore, its acceleration is zero.

(b) Below is the velocity-time graph for a body moving with uniform acceleration :

Example : The table below represents the velocity of a body at different instants, starting from rest.

Distance travelled in 8 s = S = area of triangle OPQ
= $\dfrac{1}{2}$ x base x height = $\dfrac{1}{2}$ x OQ x QP = $\dfrac{1}{2}$ x 8 x 80 = 320 m

Acceleration of body = slope of line OP
= $\dfrac{\text{PQ}}{\text{QO}}$ = $\dfrac{80 - 0}{8 - 0}$ = 10 ms^-2

Hence, distance covered is 320 m and acceleration = 10 ms^-2

(a) Below is the acceleration-time graph for a body moving with uniform velocity :

In the above figure, if the body is moving with a uniform velocity, the acceleration is zero. Hence, acceleration-time graph in such a case is a straight line coinciding with time axis.

An example of a body moving with uniform velocity is a car travelling on a straight and flat road with a constant speed of 60 km/hour (or any other constant speed).

(b) Below is the acceleration-time graph for a free falling body :

In the figure given above, the straight line AF, represents the acceleration-time graph for a body falling freely with uniform acceleration of 10 ms^-2.

For example, a body falling freely under gravity moves with a uniform acceleration of 9.8 ms^-2 (or nearly 10 ms^-2).

Velocity

Reason — As, velocity is the ratio of displacement and time, therefore, the slope of displacement-time graph gives the velocity.

(i) uniform
Reason — As the graph shows the linear relationship between displacement and time i.e., the car travels equal distance in equal intervals of time in a certain direction. Thus, it is moving with uniform motion.

(ii) 10 ms^-1
Reason — As, velocity is the ratio of displacement and time, therefore,

$\text{Velocity} = \text{Slope of straight line AC} \\[0.5em] = \dfrac{(60 - 30) \text{ m}}{(5 - 2) \text{ s}} \\[0.5em] = \dfrac{30 \text{ m}}{3 \text{ s}} \\[0.5em] = 10 \text{ m s}^{-1}$

(iii) 40 m
Reason — When we observe the graph, we notice that at t = 3 s, the displacement axis shows 40 m, hence, displacement at t = 3 s is 40 m

The correct displacement-time graph for uniform motion is :

Reason — When the displacement-time graph is a straight line inclined to the time axis, then the body covers equal distances in equal intervals of time and shows uniform motion.

Both (1) and (2)

Reason — (1) Since, velocity x time = displacement, the area enclosed between the velocity-time sketch and X-axis (i.e., the time axis) gives the displacement of the body.

(2) Since, acceleration is equal to the ratio of change in velocity and time taken, therefore the slope (or gradient) of the velocity-time sketch gives the acceleration.

acceleration is uniform

Reason — When the velocity-time graph of a body in motion is a straight line inclined to the time axis then there are equal changes in velocity in equal intervals of time and hence, the movement is with uniform acceleration.

a straight line inclined to the time axis

Reason — If the motion is uniformly retarded (i.e., its velocity decreases by an equal amount in each second), the velocity-time graph will be a straight line inclined to the time axis with a negative slope.

(i) 125 m
Reason — Area under the velocity-time graph will give the distance travelled in 5 secs :

Area = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x 5 x 50

= $\dfrac{250}{2}$

= 125 m

Hence, distance travelled in 5s = 125 m

(ii) 10 ms^-2
Reason —

$\text{Retardation} = \text{Slope of straight line} \\[0.5em] = \dfrac{(50 - 0)}{(5 - 0)} \\[0.5em] = \dfrac{50 \text{ m}}{5 \text{ s}} \\[0.5em] = 10 \text{ m s}^{-2}$

Hence, retardation = 10 ms^-2

The graph representing the state of rest of an object is :

Reason — As distance is not changing in distance-time graph hence, the state of rest of an object is shown.

The graph representing the uniform motion of an object is :

Reason — As in velocity-time graph, velocity of the body is not changing, therefore, the body is covering equal distance in equal intervals of time and is in uniform motion.

Uniform velocity

Reason — As in velocity-time graph, velocity of the body is not changing, therefore, the body is covering equal distance in equal intervals of time and is in uniform motion.

$\dfrac{1}{4}$

Reason — Distance covered in last 2 s = area of triangle = $\dfrac{1}{2}$ x 2 x 10 = 10 m

Distance covered in last 7 s = area of trapezium = $\dfrac{1}{2}$ x 10 x (2 + 6) = 40 m

Hence, ratio of of the distance travelled by the object in the last 2 s and the distance travelled in 7 s is = $\dfrac{10}{40}$ = $\dfrac{1}{4}$

8 m, 16 m

Reason — Displacement = area of 1st rectangle - area of 2nd rectangle + area of third rectangle

= [4 x 2] - [2 x 2] + [2 x 2] = 8 m

Distance = area of 1st rectangle + area of 2nd rectangle + area of third rectangle

= [4 x 2] + [2 x 2] + [2 x 2] = 16 m

Reason —

Option (1) :

Velocity graph is incorrect as it always shows positive velocity and height graph is also incorrect because it shows that height of ball was zero for some amount of time whereas in the question it is mentioned that ball was in contact with floor for negligible time.

Option (2) :

Velocity graph is incorrect because it's linear sections with sign reversal is not following height-time graph but height graph is correct as it is smooth parabolic curves representing motion to and from height h.

Option (3) :

Velocity graph is correct because it's linear sections with sign reversal is following height-time graph and height-time graph is also correct as it is smooth parabolic curves representing motion to and from height h.

Option (4) :

Velocity graph is incorrect because the process of sign reversal of velocity should be instantaneous and have no time gap during sign reversal. Height graph is also incorrect as it shows the ball has started it's motion from ground (h = 0) which is not the case as the ball is released from a particular height.

We observe from the given displacement-time graph above, that the slope is a straight line inclined with time axis, so the body is moving with uniform velocity. Hence, the velocity will be same at t = 1 s, 2 s and 3 s.

$\text{Velocity} = \text{Slope of straight line OP} \\[0.5em] = \dfrac{\text{AB}}{\text{OB}} \\[0.5em] = \dfrac{(2 - 0) \text{ m}}{(1 - 0) \text{ s}} \\[0.5em] = \dfrac{2 \text{ m}}{1 \text{ s}} \\[0.5em] = 2 \text{ m s}^{-1}$

Hence, velocity at t = 1 s, 2 s and 3 s is equal to 2 m s^-1

Velocity-time graph for the motion of a body is given below :

(a) The displacement-time sketch for the car is shown below :

As we know,

$\text {Avg. velocity of car} = \dfrac {\text {Total distance}}{\text {Total time}} \\[1em] \text {Avg. velocity of car} = \dfrac {20 - 0 \text { m}}{4 - 0\text { s}} \\[1em] \text {Avg. velocity of car} = \dfrac {20 \text { m}}{4\text { s}} \\[1em] \Rightarrow \text {Avg. velocity of car} = 5\text { ms} ^ {-1}$

Hence, average velocity of the car is 5 ms^-1.

(b) When we observe the graph, we can find the displacement of the car at various points,

At t = 2.5 s, displacement = 12.5 m and

At t = 4.5 s, displacement = 22.5 m

The displacement-time graph is given below :

(i) When we observe the graph, we find —

Total distance travelled in interval 1 s to 5 s = 18 m - 6 m = 12 m

Hence, total distance travelled in interval 1 s to 5 s = 12 m

(ii) As we know,

$\text {Avg. velocity} = \dfrac {\text {Total distance}}{\text {Total time}} \\[1em] \text {Avg. velocity} = \dfrac {(18 - 6) \text {m}}{(5 - 1)\text {s}} \\[1em] \text {Avg. velocity} = \dfrac {12 \text {m}}{4\text {s}} \\[1em] \Rightarrow \text {Avg. velocity} = 3\text {ms} ^ {-1}$

Hence, average velocity of the car in interval 1 s to 5 s is 3 ms^-1.

Below is the displacement-time graph of the body with the different points marked :

(a) As we know,

(i) At t = 0 to 5 s

Velocity = Slope of straight line OA

$\text {Velocity} = \dfrac {\text {AE}}{\text {OE}} = \dfrac {(3 - 0) \text {m}}{(5 - 0) \text {s}} \\[0.5em] \Rightarrow \text {Velocity} = \dfrac {3 \text {m}}{5 \text {s}} \\[0.5em] \Rightarrow \text {Velocity} = 0.6 {\text {ms}}^{-1} \\[0.5em]$

Hence, velocity at t = 0 to 5 s = 0.6 m s^-1

(ii) At t = 5 to 7 s

In this part we observe that there is no change in Y axis, (i.e. displacement is zero so the body is stationary).

Hence, velocity at t = 5 to 7 s = 0 m s^-1

(iii) At t = 7 s to 9 s

Velocity = Slope of straight line BC

$\text {Velocity} = \dfrac {\text {CD}}{\text {BD}} = \dfrac {(7 - 3) \text {m}}{(9 - 7) \text {s}} \\[0.5em] \Rightarrow \text {Velocity} = \dfrac {4 \text {m}}{2 \text {s}} \\[0.5em] \Rightarrow \text {Velocity} = 2 {\text {ms}}^{-1} \\[0.5em]$

Hence, velocity at t = 7 s to 9 s = 2 m s^-1

(b) As we know,

$\text {Avg. velocity} = \dfrac {\text {Total distance}}{\text {Total time}} \\[0.5em]$

Substituting the values from the graph we get,

$\text {Avg. velocity} = \dfrac {(7 - 3) \text {m}}{(9 - 5)\text {s}} \\[0.5em] \text {Avg. velocity} = \dfrac {4 \text {m}}{4\text {s}} \\[0.5em] \Rightarrow \text {Avg. velocity} = 1\text {ms} ^ {-1} \\[0.5em]$

Hence, average velocity of the car is 1 ms^-1.

(i) As we know,

$\text {Avg. velocity} = \dfrac {\text {Total distance}}{\text {Total time}}$

Substituting the values from the graph, for the first 4 s, we get,

$\text {Avg. velocity} = \dfrac {10 \text {m}}{4 \text {s}} \\[0.5em] \Rightarrow \text {Avg. velocity} = 2.5\text {ms} ^ {-1} \\[0.5em]$

Hence, average velocity of the car is 2.5 ms^-1.

(ii) From the graph, we get,

Displacement = final position - initial position

Substituting the values from the graph, for the first 10 s, we get,

Displacement = -10 - 0 = -10 m

Hence, displacement from the initial position at the end of 10 s = - 10 m

(iii) The cyclist would reach the start point two times, one at the 7 s and the other at 13 s.

(i) When we observe the graph, we find that the car B was ahead of car A by 40 km.

(ii) As we know,

$\text{Velocity} = \dfrac{\text{displacement}}{\text{time}}$

Cars A and B have uniform velocities as displacement-time graph for both are straight lines.

At t = 4 h

Car A

$\text {Velocity} = \dfrac {160 - 0 \text {km}}{4 - 0 \text {h}} \\[0.5em] \Rightarrow \text {Velocity} = \dfrac {160 \text {km}}{4 \text {h}} \\[0.5em] \Rightarrow \text {Velocity} = 40 {\text {kmh}}^{-1} \\[0.5em]$

Hence, velocity of car A = 40 km h^-1

Car B

$\text {Velocity} = \dfrac {120 - 40 \text {km}}{4 - 0 \text {h}} \\[0.5em] \Rightarrow \text {Velocity} = \dfrac {80 \text {km}}{4 \text {h}} \\[0.5em] \Rightarrow \text {Velocity} = 20 {\text {kmh}}^{-1} \\[0.5em]$

Hence, velocity of car B = 20 km h^-1

When we observe the graph, we find that both cars A and B intersect at a point K.

Hence,

(iii) the car A catches car B at time (t) = 2 h.

(iv) the distance from start when the car A will catch the car B = 80 km.

Below is the displacement-time graph of the body made to fall from the top of a tower:

When we observe the graph, we find, that the displacement-time graph is a curve.

Hence, we can say that the motion is non-uniform.

As we know, the displacement of body at any instant can be obtained by finding the area enclosed by the straight line with the time axis up to that instant.

Let displacements at t = 1 s, 2 s, 3 s, 4 s be S₁, S₂, S₃, S₄ respectively.

At t = 1 s,

S₁ = $\dfrac{1}{2}$ x 1 x 1 = 0.5 m

At t = 2 s,

S₂ = $\dfrac{1}{2}$ x 2 x 2 = 2 m

At t = 3 s,

S₃ = $\dfrac{1}{2}$ x 3 x 3 = 4.5 m

At t = 4 s,

S₄ = $\dfrac{1}{2}$ x 4 x 4 = 8 m

The table below gives the displacement of body at different instants.

The displacement time graph is shown below :

(i) The distance travelled in any part of the graph can be obtained by finding the area enclosed by the graph in that part with the time axis.

(ii) Let E be the point on time axis (x axis) at time t and F be the point at time 2t.

When we observe the graph, we find,

Distance travelled in part BC = Area of rectangle EBCF

∴ Distance travelled in part BC = length x breadth
= (2t - t) x (v₀ - 0)
= t x v₀     .... [1]

Distance travelled in part AB = Area of triangle ABT

∴ Distance travelled in part AB = $\dfrac{1}{2}$ x base x height
= $\dfrac{1}{2}$ x (t - 0) x (v₀ - 0)

= $\dfrac{1}{2}$ x t x v₀     .... [2]

Comparing [1] and [2] we get,

Distance travelled in part BC : Distance travelled in part AB

$= \dfrac {\text {t v}_0}{\dfrac{\text {tv}_0}{2}} \\[0.5em] = \dfrac {2}{1}$

Hence,
Distance travelled in part BC : Distance travelled in part AB = 2 : 1

(iii) The different parts of the graph are mentioned below:

(a) Uniform velocity is shown in part BC of the graph, as the velocity is constant with time.

(b) Uniform acceleration is shown in part AB of the graph, as the velocity is increasing with time.

(c) Uniform retardation is shown in part CD of the graph, as the velocity is decreasing with time.

(iv) (a) The magnitude of acceleration is lower, as slope of line AB is less than that of line CD.

(b) Acceleration in part AB = slope of AB

$\text {Slope of line AB} = \dfrac{\text v_0 - 0}{\text t - 0} \\[0.5em] \Rightarrow\text {Slope of line AB} = \dfrac{\text v_0}{\text t} \\[0.5em]$

Retardation in part CD = slope of CD

$\text {Slope of line CD} = \dfrac{\text v_0 - 0}{2.5\text t - 2\text t} \\[0.5em] \Rightarrow\text {Slope of line AB} = \dfrac{\text v_0}{0.5\text t} \\[0.5em]$

Magnitude of acceleration : Magnitude of retardation = Slope of line AB : Slope of line CD

$= \dfrac{\dfrac{\text v_0}{\text t}}{\dfrac{\text v_0}{0.5\text t}} \\[1em] = \dfrac{0.5}{1} \\[1em] = \dfrac{5}{10} \\[1em] = \dfrac{1}{2}$

Hence, Magnitude of acceleration : Magnitude of retardation = 1 : 2

Let E be the point at t = 4 and F be the point at t = 8 as labelled in the graph below :

(i) Acceleration in part AB = slope of AB

$\text {slope of AB} = \dfrac{(30 - 0) \text {ms}^{-1}}{(4-0) \text {s}} \\[0.5em] \text {slope of AB} = \dfrac{30 \text {ms}^{-1}}{4 \text {s}} \\[0.5em] \text {slope of AB} = 7.5 \text {ms}^{-2} \\[0.5em]$

Hence, Acceleration in part AB = 7.5 m s^-2

Acceleration in part BC = slope of BC

We observe from the graph that there is no change in velocity in part BC. Hence, Acceleration in part BC = 0 m s^-1

Acceleration in part CD = slope of CD

$\text {slope of CD} = \dfrac{(0 - 30) \text {ms}^{-1}}{(10-8) \text {s}} \\[0.5em] \text {slope of CD} = \dfrac{-30 \text {ms}^{-1}}{2 \text {s}} \\[0.5em] \text {slope of CD} = -15 \text {ms}^{-2} \\[0.5em]$

Hence, Acceleration in part CD = -15 m s^-1

(ii) Displacement in each part is as follows —

(a) Displacement of part AB = Area of triangle ABE

$= \dfrac {1}{2} \times \text {base} \times \text {height}$

Substituting the values in the formula above, we get,

$= \dfrac {1}{2} \times {(4 -0) \text {s}} \times (30 - 0)\text {m s}^{-1} \\[0.5em] = \dfrac {1}{2} \times 4 \text {s} \times 30 \text {m s}^{-1} \\[0.5em] = 30 \text {m} \\[0.5em]$

Hence,
Displacement of part AB = 60 m

(b) Displacement of part BC = Area of Square EBCF

$= \text {length} \times \text {breadth}$

Substituting the values in the formula above, we get,

$= {(8 -4) \text {s}} \times (30 - 0)\text {m s}^{-1} \\[0.5em] = 4 \times 30 \text {m} \\[0.5em] = 120 \text {m} \\[0.5em]$

Hence,
Displacement of part BC = 120 m

(c) Displacement of part CD = Area of triangle CDF

$= \dfrac {1}{2} \times \text {base} \times \text {height}$

Substituting the values in the formula above, we get,

$= \dfrac {1}{2} \times {(10 -8) \text {s}} \times (30 - 0)\text {m s}^{-1} \\[0.5em] = \dfrac {1}{2} \times 2 \times 30 \text {m} \\[0.5em] = 30 \text {m} \\[0.5em]$

Hence,
Displacement of part CD = 30 m

(iii) Total displacement = Displacement of part AB + Displacement of part BC + Displacement of part CD
= 60 + 30 + 120 = 210

Hence,
total displacement = 210 m

The velocity-time graph for the motion of ball on a smooth floor in a straight line is shown below :

Total distance travelled = distance travelled in the first 6 s + distance travelled in next 6 s.

As we know,
Distance = velocity x time

Substituting the values in the formula above we get,

Distance when the ball moves towards the wall = 10 m s^-1 x 6 s = 60 m

Distance when the ball moves away from the wall = 10 m s^-1 x 6 s = 60 m

Total distance = 60 m + 60 m = 120 m

Displacement = distance travelled in the forward direction - distance travelled while coming back
= 60 m - 60 m = 0

∴ Displacement = 0 as the ball comes back to the initial place.

(i) As we observe the graph, we find that, the nature of motion of particle is that, the particles are uniformly accelerated from 0 to 4s and then uniformly retarded from 4s to 6s.

(ii) As we know,

displacement of particles can be obtained by finding the area enclosed by the graph in that part with the time axis up to that instance.

At t = 6 s,

Displacement = area of triangle

$= \dfrac {1}{2} \times \text {base} \times \text {height}$

Substituting the values in the formula above, we get,

$= \dfrac {1}{2} \times {(6 - 0) \text {s}} \times (2 - 0)\text {m s}^{-1} \\[0.5em] = \dfrac {1}{2} \times 6 \times 2 \text {m} \\[0.5em] = 6 \text {m} \\[0.5em]$

Hence,
Displacement of particle at t = 6 s is 6 m

(iii) No, the particle does not change its direction of motion.

(iv) At t = 0 to 4 s,

Distance covered = area of triangle

$= \dfrac {1}{2} \times \text {base} \times \text {height}$

Substituting the values in the formula above, we get,

$= \dfrac {1}{2} \times {(4 - 0) \text {s}} \times (2 - 0)\text {m s}^{-1} \\[0.5em] = \dfrac {1}{2} \times 4 \text {s} \times 2 \text {m s}^{-1} \\[0.5em] = 4 \text {m} \\[0.5em]$

Hence,
Distance covered between 0 to 4 s = 4 m

At t = 4 s to 6 s,

Distance covered = area of triangle

$= \dfrac {1}{2} \times \text {base} \times \text {height}$

Substituting the values in the formula above, we get,

$= \dfrac {1}{2} \times {(6 - 4) \text {s}} \times (2 - 0)\text {m s}^{-1} \\[0.5em] = \dfrac {1}{2} \times 2 \text {s} \times 2 \text {m s}^{-1} \\[0.5em] = 2 \text {m} \\[0.5em]$

Hence,
Distance covered between 4s to 6 s = 2 m

∴ Distance covered between 0 to 4 s : Distance covered between 4 s to 6 s = 4 : 2 = 2 : 1

(v) Acceleration in part 0 s to 4 s = slope of graph

$\text {slope} = \dfrac{(2 - 0) \text {ms}^{-1}}{(4-0) \text {s}} \\[0.5em] \text {slope} = \dfrac{2 \text {ms}^{-1}}{4 \text {s}} \\[0.5em] \text {slope} = 0.5 \text {ms}^{-2} \\[0.5em]$

Hence,
Acceleration in part 0 s to 4 s = 0.5 m s^-2

As we know,

Retardation in part 4 s to 6 s = slope of graph

$\text {slope} = \dfrac{(0 - 2) \text {ms}^{-1}}{(6 - 4) \text {s}} \\[0.5em] \text {slope} = \dfrac{- 2 \text {ms}^{-1}}{2\text {s}} \\[0.5em] \text {slope} = - 1 \text {ms}^{-2} \\[0.5em]$

Acceleration = -1 ms^-2 and as we know retardation is negative acceleration

Hence, Retardation in part 4 s to 6 s = 1 m s^-2

In a displacement time graph, the time is on the X axis and the displacement of the body is on the Y axis. As, velocity is the ratio of displacement and time, therefore, the slope gives the velocity.

When slope is positive, it implies body is moving away from the starting (reference) point.

When slope is negative, it implies body is returning towards the starting (reference) point.

No, the displacement-time sketch can never be a straight line, parallel to the displacement axis because such a line would mean that the distance covered by the body in a certain direction increases without any increase in time (i.e., the velocity of the body is infinite) which is impossible.

Below is the displacement-time graph for a boy going to school with a uniform velocity :

(i) Acceleration of a body — As acceleration is equal to the ratio of change in velocity and time taken, therefore, the slope ( or gradient ) of the velocity-time graph gives the acceleration.

(ii) Distance travelled by the body in a given time — The total distance travelled by the body is obtained by the arithmetic sum of the positive displacement and negative displacement (without sign).

(iii) Displacement of the body — As the product of velocity and time gives the displacement, therefore, the area enclosed between the velocity-time sketch and X - axis (i.e., the time axis) gives the displacement of the body.

The total displacement is obtained by adding positive displacement and negative displacement numerically with proper sign.

The following can be said about the nature of motion of a body, if its displacement-time graph is as follows —

(a) A straight line parallel to time axis — shows that the body is stationary (or no motion).

(b) A straight line inclined to the time axis with an acute angle — shows the linear relationship between the displacement and time (i.e., the body travels equal distance in equal intervals of time in a certain direction). Hence, we can say the motion is away from the starting point with uniform velocity.

(c) A straight line inclined to the time axis with an obtuse angle — shows that there is motion towards the starting point with a uniform velocity.

(d) A curve — shows that the body moves with varying speed in a fixed direction. Hence, we can say that there is motion with variable velocity.

We infer from the graph, that vehicle A is moving faster than vehicle B. It is so because, the slope of line A is more than that of line B.

If a body moves with varying speed in a fixed direction i.e., with variable velocity, the displacement-time graph is not a straight line, but it is a curve. The velocity at any instant can then be obtained by finding the slope of the tangent drawn on the curve at that instant of time.

Below is the velocity-time graph for a body moving with an initial velocity u and uniform acceleration a:

Let v be the velocity at time t for this body having initial velocity u and uniform acceleration a as shown in the graph.

Distance travelled by the body in time t = Area under the velocity-time curve.

Hence,

Distance travelled = area of trapezium OABD

= $\dfrac{1}{2}$ x sum of parallel sides x height

= $\dfrac{1}{2}$ x (OA + DB) x OD

= $\dfrac{1}{2}$ x (u + v) x t

= $\dfrac{(\text{u} + \text{v})\text{t}}{2}$

(a) The graph shows uniformly accelerated motion.

Example — Uniform acceleration is depicted by the motion of a body released downwards.

(b) The graph shows the motion with a variable retardation.

Example — Motion with a variable retardation is shown by a car approaching its destination.

We can infer from the graph, that the car B has greater acceleration than car A as the slope of straight line for car B is more than that of car A.

(a) Below is the velocity-time graph for a body moving with uniform velocity :

(b) Below is the velocity-time graph for a body moving with uniform acceleration :

Retardation is calculated from the velocity-time graph, by finding the negative slope.

The following information is obtained from the given figures —

• A → Zero acceleration since slope (i.e., velocity) is constant.
• B → Zero acceleration since slope (i.e., velocity) is constant.
• C → Negative acceleration (or retardation) since slope is decreasing with time.
• D → positive acceleration since slope is increasing with time.

Below is the graph for acceleration against time for a uniformly accelerated motion:

For linear motion,

Acceleration x time = change in speed

Therefore, we can find the change in speed from the area enclosed between the acceleration-time sketch and the time axis.

Below is the velocity-time graph for the free fall of a body under gravity, starting from rest :

For a freely falling body (i.e., motion under uniform acceleration ), the displacement is directly proportional to the square of time.

                   S ∝ t²

Below is the graph of distance fallen against square of time for a freely falling body :

The slope is half the acceleration due to gravity.

Thus, the value of acceleration due to gravity (g) can be obtained by doubling the slope of the S - t² graph for a freely falling body.

For the motion with uniform velocity, distance is directly proportional to time.

As, velocity is the ratio of displacement and time, therefore, the slope gives the velocity.

When slope is positive, it implies body is moving away from the starting (reference) point.

When slope is negative, it implies body is returning towards the starting (reference) point.

For a motion in one direction in a straight line, since the direction of motion does not change, so displacement-time graph and the distance-time graph are same.

A greater slope of displacement-time graph shows higher velocity.

The negative slope of a displacement-time graph shows the motion towards the origin.

As the ratio of change in velocity and time taken is equal to the acceleration. Therefore, the slope of the velocity-time graph represents the acceleration.

On the velocity-time graph, larger the slope, higher is the acceleration or retardation.

If the velocity-time graph has a negative slope, the motion of body decreases at a constant rate (i.e., motion is uniformly retarded).

both A and R are true and R is the correct explanation of A

Explanation

Assertion (A) is true because if a body moves and comes back to its starting point (e.g., in a circular path or a round trip), the displacement is zero because the initial and final positions are the same, but the distance travelled is not zero.

Reason (R) is true because by definition, displacement is the shortest straight line distance between the initial and final positions of the body. So, the reason correctly explains the assertion.

both A and R are true and R is the correct explanation of A

Explanation

Assertion (A) is true because :

$\text {average speed} = \dfrac {\text {total distance}}{\text {total time}}$

and

$\text {average velocity} = \dfrac {\text {net displacement}}{\text {total time}}$

So if the path is not straight (e.g., in a circular or zig-zag path), distance > displacement, so average speed > average velocity in such cases.

Reason (R) is true because speed has only magnitude (no direction) i.e., scalar and velocity has both magnitude and direction i.e., vector. Hence, the reason correctly explains why average speed and average velocity can differ because velocity depends on displacement (vector) and speed depends on distance (scalar).

assertion is true but reason is false

Explanation

Assertion (A) is true because the slope of a displacement-time graph is :

$\text {Slope} = \dfrac {\text {Change in displacement}}{\text {Change in time}} = \text {velocity}$

Reason (R) is false because velocity is not the product of displacement and time.

Instead,

$\text {Velocity} = \dfrac {\text {Displacement}}{\text {Time}}$

both A and R are true and R is the correct explanation of A

Explanation

There is a misprint in the question, velocity axis should be printed as time axis.

Assertion (A) is true because a straight line parallel to the time (x) axis implies same velocity for at all times so this happens when the velocity remains constant over time, i.e., uniform velocity.

Reason (R) is true because a horizontal velocity-time graph (parallel to the time axis) indicates uniform velocity which shows the velocity doesn’t change with time and hence, Reason justifies the assertion.

both A and R are true and R is the correct explanation of A

Explanation

Assertion (A) is true because if a body moves with constant acceleration, the acceleration-time graph is indeed a straight horizontal line parallel to the time axis.

Reason (R) is true because for a freely falling body under gravity (ignoring air resistance), the acceleration is constant and equal to g (= 9.8 m/s²) and this produces a horizontal acceleration-time graph so the reason correctly explains the assertion.

assertion is false but reason is true

Explanation

Assertion (A) is false because acceleration can be positive, negative (retardation), or zero depending on whether the velocity is increasing, decreasing, or constant.

Reason (R) is true because acceleration of a body is given by :

$\text {Acceleration}= \dfrac{\text {Change in velocity}}{\text {Time interval}}$

(i) By definition,

$\text {Acceleration} = \dfrac {\text {change in velocity}}{\text {time taken}} \\[0.5em] \Rightarrow \text {Acceleration} = \dfrac {\text {final velocity - initial velocity}}{\text {time taken}} \\[0.5em]$

Hence,

$\text {a} = \dfrac{\text{v} - \text{u}}{\text {t}} \\[0.5em] \text {at} = \text{v} - \text{u} \\[0.5em] \Rightarrow \text {v} = \text{u} + \text{at} \\[0.5em]$

Hence, we get,

v = u + at     [Eq. 1]

(ii) By definition,

$\text {Distance travelled} = \text {Avg. Velocity} \times \text{time} \\[1em] \text {Distance travelled} = \Big(\dfrac{\text {Initial Vel.} + \text {Final Vel.}}{2}\Big) \times \text{time}$

or

$\text{S} = \Big(\dfrac {\text{u} + \text{v}}{2}\Big) \times \text{t}$

But, from Eq. 1, v = u + at

Therefore,

$\text{S} = \Big(\dfrac{\text {u + ( u +at)}}{2}\Big) \times \text {t} \\[0.5em] \text{S} = \Big(\dfrac{\text {2u + at}}{2}\Big) \times \text {t} \\[0.5em] \Rightarrow \text{S} = \text {ut} + \dfrac{1}{2} \text {at}^2 \\[0.5em]$

Hence, we get,
S = ut + $\bold{\dfrac{1}{2}}$ at²

(iii) By definition,

$\text {Distance travelled} = \text {Avg. Velocity} \times \text{time} \\[0.5em] \Rightarrow \text{S} = \Big(\dfrac {\text{u} + \text{v}}{2}\Big) \times \text{t}$

But, from Eq. 1, v = u + at

or t = $\dfrac{\text{v}-\text{u}}{\text{a}}$

Therefore,

$\text{S} = \Big(\dfrac {\text{v} + \text{u}}{2}\Big) \times \Big(\dfrac{\text{v}-\text{u}}{\text a}\Big) \\[1em] \Rightarrow \text{S} = \dfrac{\text{v}^2 - \text{u}^2}{2\text{a}}$

So, v² - u² = 2aS

Hence,

v² = u² + 2aS