Motion Worksheet-8

Motion Worksheet-8

 

  1. A racing car has a uniform acceleration of 4 m/s2.what distance will it cover in 10 seconds after the start ?

A. 400 metres                               B. 300 metres

C. 200 metres                               D. 100 metres

 

  1. A scooter moving at a speed of 10 m/s is stopped by applying brakes which produces a uniform acceleration   of, –0.5 m/s2. How much distance will be covered by the scooter before it stops?  

A. 50 metres                                  B. 200 metres

C. 150 metres                                D. 100 metres

 

  1. A car traveling at 20 km/h speeds up to 60 km/h in 6 seconds. What is its acceleration ?

A. 1.2 m/s2                                     B. 1.85 m/s2

C. 2.5 m/s2                                     D. 1.50 m/s2

 

  1. A bus increases its speed form 20 km/h to 50 km/h in 10 seconds. Its acceleration is :

A. 30 m/s2         B. 3 m/s2           C. 18 m/s2         D. 0.83 m/s2

 

  1. A cyclist goes around a circular track once every 2 minutes. If the radius of the circular track is 100 metres, calculate his speed. (Given π = 22/7)

A. 6.5 m/s         B. 5.5 m/s          C. 7.5 m/s                   D. 5.6 m/s

 

Answer:

  1. C

Explanation: Here, initial velocity, u = 0

          Time, t = 10 s

          Acceleration, a = 4 m/s2

And, Distance, s = ?                       

Now, putting these values in the second equation of motion :

                  

We get :   

                   s = 200 m

thus, the distance covered by the car in 10 seconds is 200 metres.

 

  1. D

Explanation: Here, initial speed, u = 10 m/s

Final speed, v = 0 (scooter stops)

Acceleration, a = –0.5 m/s2

 Distance covered, s = ?                          

Now, putting these values in the third equation of motion :

                   v2 = u2 + 2a

We get :    (0)2 = (10)2 + 2 × (–0.5) × s

                   0 = 100 – s

                   S = 100 m

Thus, the distance covered is 100 metres.

 

  1. B

Explanation: Here, initial speed, u = 20 km/h

                  

                   = 5.55 m/s                   …(1)

Final speed, v = 60 km/h

                  

                   = 16.66 m/s               …(2)

          Acceleration, a = ?         

And, Time, t = 6 s

We know that : v = u + at

So,    16.66 = 5.55 + a × 6

          6a = 16.66 – 5.55

         6a = 11.11

        

Thus, acceleration, a = 1.85 m/s2

 

Explanation: Here, initial speed, u = 20 km/h

                  

                   = 5.5 m/s                 ………(1)

Final speed, v = 50 km/h

                  

                   = 13.8 m/s                …………(2)

          Acceleration, a = ?

And, time, t = 10 s

Putting the value in the equation of motion ;

Now, v = u + at

So,    13.8 = 5.5 + a × 10

          10a = 13.8 – 5.5

          10a = 8.3

          a = (8.3/10)

          a = 0.83 m/s2

thus, the acceleration is 0.83 m/s2.

 

  1. C

Explanation: We know that for a body moving in a circular path :

          v = 2πr/t

Here, speed, v = ?               

          π = 22/7

Radius of circular track, r = 105 m

And, Time taken for 1 round, t = 2 minutes

          = 2 × 60 seconds = 120 s

Now, putting these values of π, r and t in the above formula, we get :

          = 5.5 m/s

Thus, the speed of cyclist on the circular track is 5.5 metres per second.