Motion : NCERT Exercise Questions



Q.1     An 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?
Sol.    
Here we have,
           Diameter = 200 m, therefore, radius = 200m/2 = 100 m
           Time of one rotation = 40s
           Time after 2m20s = 2 x 60s + 20s = 140s
           Distance after 140 s = ?
           Displacement after 140s =?

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                                                                  Circular track with diameter of 200 m


           We know that, velocity along a circular path = {{Circum\,ference} \over {time}}
            \Rightarrow v = {{2\pi r} \over {40s}}
            \Rightarrow v = {{2 \times 3.14 \times 100m} \over {40s}}
            \Rightarrow v = {{628m} \over {40s}} = 15.7m/s
           (a) Distance after 140s
           We know that,distance=velocity ×time
           ⇒distance=15.7m/s ×140 s = 2198 m

           (b) Displacement after 2 m 20 s i.e. in 140 s
           Since,rotatin in 40 s = 1
           Therefore, rotation in 1 s = {1 \over {40}}
           Therefore, rotation i n 140s = {1 \over {40}} \times 140 = 3.5
           Therefore, in 3.5 rotations athlete will be just at the opposite side of the circular track, i.e. at a distance equal to the diameter of the circular track which is equal to 200 m.
           Therefore,
           Distance covered in 2 m 20 s = 2198 m
           And, displacement after 2 m 20 s = 200m


Q.2     Joseph 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?
Sol.
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           Here we have,
           Distance from point A to B = 300 m
           Time taken = 2 minute 30 second = 2 x 60 + 30 s = 150 s
           Distance from point B to C = 100 m
           Time taken = 1 minute = 60 s

           (a) Average speed and velocity from point A to B
           We know that average speed = {{Total\,\,dis\tan ce} \over {Time\,\,taken}}
            \Rightarrow Average\,\,speed = {{300m} \over {150s}} = 2m/s
           Therefore, velocity = 2m/s east

           (b) Average speed and velocity from B to C
           We know that average speed = {{Total\,\,dis\tan ce} \over {Time\,\,taken}}
            \Rightarrow Average\,\,Speed = {{100m} \over {60s}} = 1.66m/s
           Therefore,average velocity=1.66 m/s west


Q.3     Abdul, while driving to school, computes the average speed for his trip to be 20 km/h. On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed for Abdul’s trip?
Sol.     Strategy: We need to calculate the time taken in each of the trip. After that, we can calculate the average speed.
           Let the distance of the school = s km
           Let time to reach the school in first trip = t1
           Let time to reach the school in second trip = t2
           We know that, Average speed = {{Total\,\,dis\tan ce} \over {Total\,\,time\,\,taken}}
           Therefore, Average speed in first trip = {s \over {{t_1}}}
            \Rightarrow 20kg/h = {s \over {{t_1}}}       {t_1} = {s \over {20}}h
           Therefore, Average speed in second trip = {s \over {{t_2}}}
            \Rightarrow 30kg/h = {s \over {{t_2}}}    {t_2} = {s \over {30}}h
           Now, total time \left( {{t_1} + {t_2}} \right) = {s \over {20}} + {s \over {30}}
            \Rightarrow \left( {{t_1} + {t_2}} \right) = {{3s + 2s} \over {60}}h = {{5s} \over {60}}h = {s \over {12}}h
           Now, average speed while both of trip = {{Total\,\,dis\tan ce\,\,{\mathop{\rm cov}} ered} \over {Total\,\,time\,\,taken}}
           = {{2s} \over {{s \over {12}}}}km/h    =  {{2s \times 12} \over s}km/h = 24km/h
           Therefore, average speed of Adbul = 24 km/h
           Therefore, average speed of Adbul = 24 km/h


Q.4     A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s2 for 8.0 s. How far does the boat travel during this time?
Sol.     Here we have,
           Initial velocity (u) = 0
           Acceleration (a) = 3.0m/s2
           Time = 8 s
           Therefore, distance (s) covered =?
           We know that, s = ut + {1 \over 2}a{t^2}
            \Rightarrow s = 0 \times 8 + {1 \over 2}3m/{s^2} \times {\left( {8s} \right)^2}
            \Rightarrow s = {1 \over 2} \times 3 \times 64m
            \Rightarrow s = 3 \times 32m
            \Rightarrow s = 3 \times 32m
            \Rightarrow s = 96m
           Therefore, boat travel a distance of 96 m in the given time
   


Q.5     A driver of a car travelling at 52 km/h applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km/h in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
Sol.     Given for first driver,
           Initial velocity, u = 52 km {h^{ - 1}} = {{52 \times 1000m} \over {60 \times 60s}} = 14.4m{s^{ - 1}}
           Time, t = 5s
           Final velocity, v = 0 (Since car stops)
           Therefore, distance,s =?
           Given for second driver, 
           Initial velocity, u = 3km {h^{ - 1}} = {{3000m} \over {60 \times 60s}} = 9.4m{s^{ - 1}}
           Time, t = 10s
           Final velocity,v = 0 
 

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           In the graph, blue slope shows the velocity of the first car and green slope shows the velocity of the second car.
           Distance is calculated by the area under the slope of the graph.
           Thus, distance covered by 1st car = Area of \Delta OAD
            \Rightarrow Distance, s = {1 \over 2} \times OD \times OA
            \Rightarrow s = {1 \over 2} \times 14.4m/s \times 5s = 7.2m/s \times 5s = 36m
           Thus, distance covered by 2nd car = Area of \Delta OBC
            \Rightarrow Distance, s = {1 \over 2} \times OC \times OB
            \Rightarrow s = {1 \over 2} \times 9.4m/s \times 10s = 4.7m/s \times 10s = 47m
           Therefore, 2nd car travelled farther


Q.6     Shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions:

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           (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?
Sol.
     (a) It is clear from graph that B covers more distance in less time. Therefore, B is the fastest.
           (b) All of them never come at the same point at the same time.
           (c) According to graph; each small division shows about 0.57 km.
                A is passing B at point S which is in line with point P (on the distance axis) and shows about 9.14 km
                Thus, at this point C travels about 9.14 – (0.57 x 3.75) km = 9.14 km – 2.1375 km = 7.0025 km ≈ 7 km
                Thus, when A passes B, C travels about 7 km.
           (d) B passes C at point Q at the distance axis which is ≈ 4km + 0.57km x 2.25 = 5.28 km
                Therefore, B traveled about 5.28 km when passes to C.


Q.7     A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m/s2, with what velocity will it strike the ground? After what time will it strike the ground?
Sol.     Here we have,
           Initial velocity,u=0
           Distance,s=20m
           Acceleration,a= 10 m s-2
           Final velocity,v=?
           Time,t = ?
           (a) Calculation of Final velocity, v
           We know that, {v^2} = {u^2} + 2as
            \Rightarrow {v^2} = 0 + 2 \times 10m/{s^2} \times 20m
            \Rightarrow {v^2} = 400{m^2}{s^{ - 2}}
            \Rightarrow v = \sqrt {400{m^2}{s^{ - 2}}}
            \Rightarrow v = 20m{s^{ - 1}}

           (b) Calculation of time, t
           We know that, v = u + at
            \Rightarrow 20m{s^{ - 1}} = 0 + 10m{s^{ - 2}} \times t
            \Rightarrow t = {{20m{s^{ - 1}}} \over {10m{s^{ - 2}}}} = 2s
           Therefore,
           Ball will strike the ground at the velocity of 20 {m{s^{ - 1}}}
           Time taken to reach at the ground = 2s 


Q.8     The speed-time graph for a car is shown is.
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           (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?

Sol.50

          (a) Distance travelled by car in the 4 second
          The area under the slope of the speed – time graph gives the distance travelled by an object.
          In the given graph
          56 full squares and 12 half squares come under the area slope for the time of 4 second.
          Total number of squares = 56 + 12/2 = 62 squares
          The total area of the squares will give the distance travelled by the car in 4 second.
          On the time axis, 5 squares = 2s
          Therefore, 1 square = {2 \over 5}s
          On the speed axis 3 squares = 2m/s
          Therefore, 1 square = {2 \over 3}m/s
          Thus, area of 1 square = {2 \over 5}s \times {2 \over 3}m/s = {4 \over {15}}m
          Therfore, area of 62 squares = {4 \over {15}}m \times 62
          = {{248} \over {15}}m = 16.53m
          Therefore, car travels 16.53 m in first 4 second.
          (b) Part MN of the slope of the graph is straight line parallel to the time axis, thus this potion of graph represents uniform motion of car.


Q.9     State 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 in a certain direction with an acceleration in the perpendicular direction.
Sol.     (a) The term acceleration implies that the velocity of the object is changing; inspite of that constant acceleration with zero velocity is impossible. When an object is thrown in upward direction, at the maximum height the velocity of the object becomes zero but still in that condition a constant acceleration due to gravity is working.

           (b) Object moving in a certain direction with an acceleration in perpendicular direction is possible; in case of circular motion. When an object moves on a circular path, its direction is along the tangent of the circle but acceleration is towards the radius of the circle. We know, that a tangent always makes a right angle with the radius; so when an object is in circular motion, the acceleration and velocity are in mutually perpendicular direction.


Q.10     An 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.
Sol.       Here we have,
             Radius, r = 42250km
             Time, t = 24 hours
             Speed =?
            We know that velocity along a circular path = {{2\pi r} \over {time}}
             \Rightarrow v = {{2 \times {{22} \over 7} \times 42250km} \over {24h}}
             \Rightarrow v = {{2 \times 22 \times 42250} \over {7 \times 24}}km/h
             \Rightarrow v = 11065.47\,km/h
            Thus, speed of the given staelline = 11065.47 km/h

 



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