Table of Contents

Numerical Problems

 

 

2.1 A train moves with a uniform velocity of 36 kmh-1 for 10 s. Find the distance travelled by it.                                                                                                                                                     (100 m)

Solution:      Velocity = V = 36 kmh-1 = 36 x 1000/ 60 x 60 = 36000/ 3600 = 10 ms-1

                       Time t = 10 s

Distance = S =?

S = Vt

S = 10 x 10 = 100 m

 

 

 

2.2 A train starts from rest. It moves through 1 km in 100 s with uniform acceleration. What will be its speed at the end of 100 s.?   (20 ms-1)

Solution:         Initial velocity Vi = 0 ms-1

Distance S = 1 km = 1000 m

Time = 100 s

Final velocity Vf =?

S = Vi x t +1/2 x a x t2

1000 = 0 x 100 + 1/2 x a x (100)2

1000 = 1/2 x 10000a

1000 = 5000a

a = 1000/5000 = 0.2 ms-2

Now using 1st equation of motion

Vf = Vi + at

Vf = 0 + 0.2 x 100

Vf = 20 ms-1

 

 

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2.3 A car has a velocity of 10 ms-1. It accelerates at 0.2 ms-2 for half minute. Find the distance travelled during this time and the final velocity of the car. (390 m, 16 ms-1)

Solution:           Initial velocity = Vi = 10 ms-1

Acceleration a = 0.2 ms-2

Time t = 0.5 min = 0.5 x 60 = 30 s

  • Distance S =?
  • Final velocity Vf =?

S = Vi x t + 1/2 x a x t2

S= 10 x 30 + 1/2 x 0.2 x (30)2

S = 300 + 1/2 x 0.2 x 90

S = 300 + 1/2 x 2/10 x 90

S = 300 + 90

S = 390 m

  • Using 1st equation of motion

Vf = Vi + at

Vf = 10 + 0.2 x 30

Vf = 10 + 6

Vf = 16 ms-1

 

 

 

2.4 A tennis ball is hit vertically upward with a velocity of 30 ms-1, it takes 3 s to reach the highest point. Calculate the maximum height reached by the ball. How long it will take to return to the ground? (45 m, 6 s)

Solution:        Initial velocity = Vi = 30 ms-1

Acceleration due to gravity g = -10 ms-2

Time to reach maximum height = t = 3 s

Final velocity Vf = 0 ms-1

                      (I)             Maximum height attained by the ball S =?

                      (II)            Time taken to return to ground t =?

S = Vi x t + 1/2 x g x t2

S = 30 x 3 + 1/2 x (-10) x (3)2

S = 90 – 5 x 9

S = 90 – 45

S = 45 m

Total time = time to reach maximum height + time to return to the ground

= 3 s + 3 s = 6 s

 

 

 

2.5 A car moves with a uniform velocity of 40 ms-1 for 5 s. It comes to rest in the next 10 s with uniform deceleration.

Find:

  • deceleration

  • total distance travelled by car.

                       (-4 ms-2, 400 m)

Solution:            Initial velocity = Vi = 40 ms-1

Time = t = 5 s

Final velocity = Vf = 0 ms-1

Time = 10 s

  • deceleration a =?
  • total distance S =?

Vf = Vi + at

Or                      at = Vf – Vi

                                         a = Vf – Vi/t

a = 0 – 40/10

a = -4 ms-2

Total distance travelled = S = S1 + S2

By using this relation

S1 = Vt

S1 = 40 x 5

S1 = 200 m ………………………………. (i)

Now by using 3rd equation of motion

2aS2 = Vf2 – Vi2

S2 = Vf2 – Vi2/2a

S2 = (0)2 – (40)2/2 x (-4)

S2 = -1600/-8

S2 = 200 m ……………………………………… (ii)

From (i) and (ii) we get;

S = S1 + S2

Or             S = 200 m + 200 m

S = 400 m

 

 

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2.6 A train starts from rest with an acceleration of 0.5 ms-2. Find its speed in kmh-1, when it has moved through 100 m.   (36 kmh-1)

Solution:         Initial velocity Vi = 0 ms-1

Acceleration a = 0.5 ms-2

Distance S = 100 m

Final velocity Vf =?

2aS = Vf2 – Vi2

                                    2 x 0.5 x 100 = Vf2 – 0

Or            , 100 = Vf2

                                    Or              Vf2 = 100 ms-1……………………..(I)

Speed in kmh-1:

From (I) we get;

Vf = 10 x 3600/1000 = 36 kmh-1

 

 

 

2.7 A train starting from rest, accelerates uniformly and attains a velocity of 48 kmh-1 in 2 minutes. It travels at this speed for 5 minutes. Finally, it moves with uniform retardation and is stopped after 3 minutes. Find the total distance travelled by train.

Solution:  Case – I:                                                                           (6000 m)

Initial velocity = Vi = 0ms-1

Time = t = 2 minutes = 2x 60 = 120 s

Final velocity = Vf = 48 kmh-1 = 48 x 1000/3600 = 13.333 ms-1

S1 = Vav x t

S1 = (Vf + Vi)/2 x t

S1 = 13.333 + 0/2 x 120

S1 = 6.6665 x 120

S1 = 799.99 m = 800 m

 

Case – II:

Uniform velocity = Vf = 13.333 ms-1

Time = t = 5 minutes = 5 x 60 = 300 s

S2 = v x t

S2 = 13.333 x 300

S2 = 3999.9 = 4000 m

 

Case – III:

Initial velocity = Vf = 13.333 ms-1

Final velocity = Vi = 0 ms-1

Time = t = 3 minutes = 3 x 60 = 180 s

S3 = Vav x t

S3 = (Vf + Vi)/2 x 180

S3 = 13.333 + 0/2 x 180

S3 = 6.6665 x 180

S3 = 1199.97 = 1200 m

Total distance = S = S1 + S2 + S3

S = 800 + 4000 + 1200

S = 6000 m

 

 

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2.8 A cricket ball is hit vertically upwards and returns to ground 6 s later. Calculate

(i)       Maximum height reached by the ball

(ii)      initial velocity of the ball                                                   (45m, 30 ms-1)

Solution:     Acceleration due to gravity = g = -10 ms-1 (for upward motion)

Time to reach maximum height (one sided time) = t = 6/2 = 3 s

Velocity at maximum height = Vf = 0 ms-1

  • Maximum height reached by the ball S = h =?
  • The maximum initial velocity of the ball = Vi =?

Since,      Vf = Vi + gxt

Vi = Vf – gxt

Vi = 0 – (-10) x 3

                              Vi = 30 ms-1

Now using 3rd equation of motion

2aS = Vf2 – Vi2

S =       Vf2 – Vi2/2a

S = (0)2 – (30)2/2 x (-10)

S = -90/-20

S = 45 m

 

 

 

2.9 When brakes are applied, the speed of a train decreases from 96 kmh-1 to 48 kmh-1 in 800 m. How much further will the train move before coming to rest? (Assuming the retardation to be constant).                                                                                                (266.66 m)

 

Solution:      Initial velocity = Vi = 96 kmh-1 = 96 x 1000/3600 = 96000/3600 ms-1

Final velocity = Vf = 48 kmh-1 = 48 x 1000/3600 = 48000/3600 ms-1

Distance = S = 800 m

Further Distance = S1 =?

First of all, we will find the value of acceleration a

2aS = Vf2 – Vi2

2 x a x 800 = (48000/3600)2 – (96000/3600)2

1600a = (48000/3600)2 – ((2 x 48000)/3600)2        (96000 = 2 x 48000)

1600a = (48000/3600)2 ((1)2 – (2)2)        (taking (48000/36000) as common)

1600a = (48000/3600)2 (1 – 4)

1600a = (48000/3600)2 (-3)

a = (48000/3600)2 x 3/1600

now, we will find the value of further distance S2:

Vf = 0              ,           S2 =?

2aS = Vf2 – Vi2

-2 (48000/36000)2 x 3/1600 x S1 = (0)2 – (48000/36000)

S1 = (48000/36000)2 x (48000/36000)2 x 1600/3 x 2

S1 = 1600/6

S2 = 266.66m

 

 

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2.10 In the above problem, find the time taken by the train to stop after the application of brakes.                                                                                                                              (80 s)

Solution:     by taking the data from problem 2.9:

Initial velocity = Vi = 96 kmh-1 = 96 x 1000/3600 = 96000/3600 ms-1

Final velocity = Vf = 0 ms-1

a = -(48000/3600)2 x 3/1600 ms-2

time = t =?

Or   Vf = Vi + at

Or   at = Vf – Vi

t = Vf – Vi/a

t = 0 – (48000/3600)/-( (48000/3600) x 1/1600

t = – 96000/3600 x (3600/48000)2 x 1600/3

t = 2 x 48000/3600 x (3600/48000 x 3600/48000) x 1600/3

t = 2 x 3600/3 x 3

t = 2 x 40 = 80 s