Motion in One Dimension — Question 16

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.