Question 1

Question 1An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Accepted Answer

Given,Diameter of track (d) = 200 mCircumference of the track = πd = 200π mSo, distance covered in 40 seconds = 200π mDistance covered in 1 sec = 200π40\dfrac{200π}{40}40200π​Distance covered in 2 min 20 sec (i.e., 60 + 60 + 20 = 140 sec) = 200π40\dfrac{200π}{40}40200π​ x 140= 200×22×1407×40\dfrac{200 	imes 22 	imes 140}{7 	imes 40}7×40200×22×140​= 2200 mNumber of rounds completed in 140 sec = 14040\dfrac{140}{40}40140​ = 3.5∴ the final position (with respect to the initial position) is at the opposite end of the circular track. So, the net displacement will be equal to the diameter of the track = 200 m.Hence, the net distance covered by the athlete is 2200 m and the total displacement of the athlete is 200 m.

Question 2

Question 2Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph's average speeds and velocities in jogging(a) from A to B and(b) from A to C?

Accepted Answer

(a) From A to BGiven,Distance covered from A to B = 300 mTime taken to travel from A to B = 2 min and 30 sec = 60 + 60 + 30 = 150 secDisplacement from A to B = 300 mAverage speed = total distance travelledtotal time taken\dfrac{	ext{total distance travelled}}{	ext{total time taken}}total time takentotal distance travelled​Average velocity = total displacementtotal time taken\dfrac{	ext{total displacement}}{	ext{total time taken}}total time takentotal displacement​Substituting we get,Average speed = 300150\dfrac{300}{150}150300​ = 2 ms-1Average velocity = 300150\dfrac{300}{150}150300​ = 2 ms-1Hence, average speed from A to B is 2 ms-1 and average velocity = 2 ms-1(b) From A to CDistance covered from A to C = 300 m + 100 m = 400 mTime taken from A to C = 2 min 30 sec + 1 min = 60 + 60 + 30 + 60 = 210 secDisplacement from A to C = 300 m – 100 m = 200 mAverage speed = total distance travelledtotal time taken\dfrac{	ext{total distance travelled}}{	ext{total time taken}}total time takentotal distance travelled​Average velocity = total displacementtotal time taken\dfrac{	ext{total displacement}}{	ext{total time taken}}total time takentotal displacement​Substituting we get,Average speed = 400210\dfrac{400}{210}210400​ = 1.9 ms-1Average velocity = 200210\dfrac{200}{210}210200​ = 0.952 ms-1Hence, average speed from A to C is 1.9 ms-1 and average velocity = 0.952 ms-1

Question 3

Question 3Abdul, while driving to school, computes the average speed for his trip to be 20 kmh-1. On his return trip along the same route, there is less traffic and the average speed is 30 kmh-1. What is the average speed for Abdul's trip?

Accepted Answer

Distance travelled to reach the school = distance travelled to reach home = dTime taken to reach school = t1Time taken to reach home = t2Average speed = total distance travelledtotal time taken\dfrac{	ext{total distance travelled}}{	ext{total time taken}}total time takentotal distance travelled​On going to school :20 = dt1\dfrac{	ext{d}}{	ext{t}_1}t1​d​ ort1 = d20\dfrac{	ext{d}}{20}20d​On return trip :30 = dt2\dfrac{	ext{d}}{	ext{t}_2}t2​d​ ort2 = d30\dfrac{	ext{d}}{30}30d​Now, the average speed for the entire trip= d+dt1+t2\dfrac{	ext{d+d}}{	ext{t}_1 + 	ext{t}_2}t1​+t2​d+d​= 2dd20+d30\dfrac{	ext{2d}}{\dfrac{	ext{d}}{20} + \dfrac{	ext{d}}{30}}20d​+30d​2d​= 2d3d + 2d60\dfrac{	ext{2d}}{\dfrac{	ext{3d + 2d}}{60}}603d + 2d​2d​= 2d×605d\dfrac{	ext{2d} 	imes {60}}{5	ext{d}}5d2d×60​= 24 kmh-1∴ Abdul's average speed for the entire trip is 24 kmh-1.

Question 4

Question 4A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms-2 for 8.0 s. How far does the boat travel during this time?

Accepted Answer

Given,Initial velocity (u) = 0Acceleration (a) = 3 ms-2Time period (t) = 8 sDistance travelled (S) = ?According to the second equation of motion,S = ut + 12\dfrac{1}{2}21​at2Substituting we get,S = 0 + 12\dfrac{1}{2}21​ x 3 x 8 x 8= 96 m∴ The motorboat travels a distance of 96 m

Question 5

Question 5A driver of a car travelling at 52 kmh-1 applies the brakes. Shade the area on the graph that represents the distance travelled by the car during the period.(b) Which part of the graph represents uniform motion of the car?

Accepted Answer

(a) The initial speed of the graph is 52 kmh-1. After the driver applies the brakes, the speed becomes zero.The x-axis of the graph represents the time and the y-axis of the graph represents the velocity.In the graph, point A represents the initial velocity of the car, that is, 52 kmh-1, and point B represents the time when the velocity becomes zero.The distance travelled by an object is equal to the product of velocity and time.Therefore, the area under the velocity-time graph represents the distance travelled by the car during the period.Distance travelled by the car = Area of triangle AOB(b) The slope of the velocity-time graph is the same throughout. So, the whole graph represents the uniform motion of the car.

Question 6

Question 6Fig below shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:(a) Which of the three is travelling the fastest?(b) Are all three ever at the same point on the road?(c) How far has C travelled when B passes A?(d) How far has B travelled by the time it passes C?

Accepted Answer

(a) As the slope of line B is the greatest, hence, B is travelling at the fastest speed.(b) No, as the three lines do not intersect at a single point.(c) As there are 7 unit areas of the graph between 0 and 4 on the Y axis, 1 graph unit equals 47\dfrac{4}{7}74​ km.As the initial point of object C is 4 graph units away from the origin, its initial distance from the origin is 4 x 47\dfrac{4}{7}74​ km = 167\dfrac{16}{7}716​ kmWhen B passes A, the distance between the origin and C is 8 km.∴ total distance travelled by C in this time = 8 – 167\dfrac{16}{7}716​ km = 5.71 km(d) Total distance travelled by B when it crosses C = 9 x 47\dfrac{4}{7}74​ = 5.14 km

Question 7

Question 7A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms-2, with what velocity will it strike the ground? After what time will it strike the ground?

Accepted Answer

Given,Initial velocity (u) = 0Distance travelled (s) = 20 mAcceleration (a) = 10 ms-2As per the third law of motion,v2 – u2 = 2asSubstituting we get,v2 - 0 = 2 x 10 x 20⇒ v2 = 400⇒ v = 400\sqrt{400}400​⇒ v = 20 ms-1Hence, ball strikes the ground with a velocity of 20 ms-1As per the first equation of motion,v = u + atort = v - ua\dfrac{	ext{v - u}}{	ext{a}}av - u​Substituting we get,t = 20−010\dfrac{20-0}{10}1020−0​ = 2 seconds∴ the ball reaches the ground after 2 seconds.

Question 8

Question 8The speed-time graph for a car is shown in Fig. below(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.(b) Which part of the graph represents uniform motion of the car?

Accepted Answer

(a) Distance travelled by the car in the first 4 seconds = area under the slope of the speed-time graph.To find this area, we will use the square counting method.On time axis (i.e., x axis) :5 squares = 2 units∴ 1 square = 25\dfrac{2}{5}52​ unitsOn speed axis (i.e., y axis) :3 squares = 2 units∴ 1 square = 23\dfrac{2}{3}32​ unitsArea of each square = 25\dfrac{2}{5}52​ x 23\dfrac{2}{3}32​ = 415\dfrac{4}{15}154​ sq units.We have fully filled squares, more than half filled squares, half filled squares and less than half filled squares in this area.Area of fully filled square = Area of 1 square = 415\dfrac{4}{15}154​ sq unitsArea of more than half filled square = Area of 1 square = 415\dfrac{4}{15}154​ sq unitsArea of half filled square = 12\dfrac{1}{2}21​ x Area of 1 square = 12×415\dfrac{1}{2} 	imes \dfrac{4}{15}21​×154​ = 215\dfrac{2}{15}152​ sq unitsArea of less than half filled square = 0No. of fully filled squares = 56 [Marked in graph with 'F']Area of fully filled squares = 56 x 415\dfrac{4}{15}154​ = 14.93 sq unitsNo. of more than half filled squares = 4 [Marked in graph with 'M']Area of more than half filled squares = 4 x 415\dfrac{4}{15}154​ = 1.06 sq unitsNo. of half filled squares = 3 [Marked in graph with 'H']Area of half filled squares = 3 x 415\dfrac{4}{15}154​ = 0.8 sq unitsNo. of less than half filled squares = 4 [Marked in graph with 'L']Area of less than half filled squares = 4 x 0 = 0 sq units∴ Total Area = 14.93 + 1.06 + 0.4 = 16.39 sq units∴ Total distance covered in first 4 seconds = Area under speed-time graph = 16.4 mHence, the car travels 16.4 m in first 4 seconds(b) From graph, the speed of the car does not change from the points (x = 6) to (x = 10). Hence, the car is said to be in uniform motion from the 6th to the 10th second.

Question 9

Question 9State which of the following situations are possible and give an example for each of these:(a) an object with a constant acceleration but with zero velocity(b) an object moving with an acceleration but with uniform speed.(c) an object moving in a certain direction with an acceleration in the perpendicular direction.

Accepted Answer

(a) It is possible;When an object is thrown up vertically into the air, it has constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.(b) It is possible;When an object moves on a circular path with uniform speed it is still under acceleration because the velocity changes due to continuous change in the direction of motion.(c) It is possible;When an object is accelerating in a circular trajectory, the acceleration is in the perpendicular direction.

Question 10

Question 10An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Accepted Answer

Given,Radius of the orbit (r) = 42250 km∴ circumference of the orbit= 2 x 227\dfrac{22}{7}722​ x r= 2 x 227\dfrac{22}{7}722​ x 42250= 265571.42 kmTime taken to complete one revolution in the orbit = 24 hoursConvert hours into sec1 hour = 3600 secSo, 24 hour = 24 x 3600 secSpeed = DistanceTime\dfrac{	ext{Distance}}{	ext{Time}}TimeDistance​Substituting we get,Speed = 265571.4224×3600\dfrac{265571.42}{24 	imes 3600}24×3600265571.42​ = 3.07 kms-1∴ Speed of the artificial satellite = 3.07 kms-1.

Question 11

Question 1An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Accepted Answer

Yes, an object can have zero displacement if it returns to its initial position.Example: When a racer runs on a running track, his initial and final position is the same. Hence, the displacement is zero.

Question 12

Question 2A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Accepted Answer

Given,Time period of travel = 2 minutes 20 secondsConvert min into sec1 min = 60 secSo 2 min 20 sec = (2 x 60) sec + 20 sec = 120 + 20 = 140 secSide of the given square field = 10 mHence, the perimeter of a square = 4 x side = 4 x 10 = 40 mTime taken by the farmer to cover the boundary of 40 m = 40 sSo, in 1 s, the farmer covers a distance of = 4040\dfrac{40}{40}4040​ x 1 = 1 mHence, distance covered by the farmer in 140 sec = 140 mThe total number of rotations taken by the farmer to cover a distance of 140 m= total distanceperimeter\dfrac{\text{total distance}}{\text{perimeter}}perimetertotal distance​= 14040\dfrac{140}{40}40140​= 3.5If the farmer starts from point A, after 3.5 rounds he will be at point C of the field.∴ Displacement (from Pythagoras theorem)AC=(AB)2+(BC)2=(10)2+(10)2=100+100=200=102=10×1.414=14.14 m\text{AC} = \sqrt{(\text{AB})^2 + (\text{BC})^2} \[1em] = \sqrt{(10)^2 + (10)^2} \[1em] = \sqrt{100 + 100} \[1em] = \sqrt{200} \[1em] = 10\sqrt{2} \[1em] = 10 \times 1.414 \[1em] = 14.14 \text{ m}AC=(AB)2+(BC)2​=(10)2+(10)2​=100+100​=200​=102​=10×1.414=14.14 mHence, the magnitude of displacement = 14.14 m

Question 13

Question 3Which of the following is true for displacement?(a) It cannot be zero.(b) Its magnitude is greater than the distance travelled by the object.

Accepted Answer

(a) False.Reason — The displacement of an object which travels a certain distance and comes back to its initial position is zero.(b) False.Reason — The displacement of an object can be equal to, but never greater than the distance travelled.

Question 14

Question 1Distinguish between speed and velocity.

Accepted Answer

SpeedVelocityThe distance travelled per second by a moving object is called its speed.The distance travelled per second by a moving object in a particular direction is called its velocity.It is a scalar quantity. The speed does not tell us the direction of motion.It is a vector quantity. The velocity tells us the speed as well as the direction of motion.The speed is always positive since direction is not taken into consideration.The velocity can be positive or negative depending upon the direction of motion.After one round in a circular path, the average speed is not zero.After completing each round in a circular path, the average velocity is zero.

Question 15

Question 2Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Accepted Answer

As average speed is the total distance travelled in a time frame and average velocity is the total displacement in a time frame, so the magnitude of average velocity and average speed will be the equal when the total distance travelled is equal to the displacement.This happens when an object moves along a straight line in the same direction (without turning back).

Question 16

Question 3What does the odometer of an automobile measure?

Accepted Answer

An odometer is a device that measures the distance travelled by an automobile based on the perimeter of the wheel as the wheel rotates.

Question 17

Question 4What does the path of an object look like when it is in uniform motion?

Accepted Answer

The path of an object in uniform motion is a straight line.

Question 18

Question 5During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108 ms-1.

Accepted Answer

Given,Time taken (t) = 5 minConvert min into sec1 min = 60 secSo, 5 min = 5 x 60 sec = 300 secSpeed of the signal = 3 × 108 ms-1We know, speed = distancetime\dfrac{	ext{distance}}{	ext{time}}timedistance​Substituting the values, we get,3 × 108 = distance300\dfrac{	ext{distance}}{300}300distance​Distance covered (d) = 3 × 108 x 300 = 9 x 1010 m.Hence, the distance of the spaceship from the ground station = 9 x 1010 m

Question 19

Question 1When will you say a body is in(i) uniform acceleration?(ii) non-uniform acceleration?

Accepted Answer

(i) The acceleration is said to be uniform (or constant) when equal changes in velocity take place in equal intervals of time.Example — The motion of a body under gravity (e.g., free fall of a body)(ii) If changes in velocity are not same in the same intervals of time, the acceleration is said to be non-uniform.Example — The motion of a vehicle on a crowded road.

Question 20

Question 2A bus decreases its speed from 80 kmh-1 to 60 kmh-1 in 5 s. Find the acceleration of the bus.

Accepted Answer

Given,Initial velocity (u) = 80 kmh-1Convert kmh-1 to ms-1 : multiply by 518\dfrac{5}{18}185​80 x 518\dfrac{5}{18}185​ = 22.22 ms-1The final velocity (v) = 60 kmh-160 x 518\dfrac{5}{18}185​ = 16.66 ms-1Time (t) = 5 secAccording to the first equation of motion,v = u + atora = v - ut\dfrac{	ext {v - u}}{	ext{t}}tv - u​Substituting we get,a = 16.66−22.225\dfrac{16.66 - 22.22}{5}516.66−22.22​ = -1.112 ms-2∴ the acceleration of the bus is -1.112 ms-2.The negative sign indicates that the velocity of the bus is decreasing.

Motion

Exercises

Intext Questions 1

Intext Questions 2

Intext Questions 3

Intext Questions 4

Intext Questions 5