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 V_i = 0 ms^-1

Distance S = 1 km = 1000 m

Time = 100 s

Final velocity V_f =?

S = V_i 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 1^st equation of motion

V_f = V_i + at

V_f = 0 + 0.2 x 100

V_f = 20 ms^-1

 

 

 

 

 

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 = V_{i =} 10 ms^-1

Acceleration a = 0.2 ms^-2

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

• Distance S =?
• Final velocity V_f =?

S = V_i 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 1^st equation of motion

V_f = V_i + at

V_f = 10 + 0.2 x 30

V_f = 10 + 6

V_f = 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 = V_{i =} 30 ms^-1

Acceleration due to gravity g = -10 ms^-2

Time to reach maximum height = t = 3 s

Final velocity V_f = 0 ms^-1

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

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

S = V_i 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 =?

V_f = V_i + at

Or                      at = V_{f –} V_i

                                         a = V_{f –} V_i/t

a = 0 – 40/10

a = -4 ms^-2

Total distance travelled = S = S₁ + S₂

By using this relation

S₁ = V_t

S₁ = 40 x 5

S₁ = 200 m ………………………………. (i)

Now by using 3^rd equation of motion

2aS₂ = V_f² – V_i²

S₂ = V_f² – V_i²/2a

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

 

 

 

 

 

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 V_i = 0 ms^-1

Acceleration a = 0.5 ms^-2

Distance S = 100 m

Final velocity V_f =?

2aS = Vf² – V_i²

                                    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;

V_f = 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 = V_i = 0ms^-1

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

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

S₁ = V_av x t

S₁ = (V_f + V_i)/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 = V_f = 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 = V_f = 13.333 ms^-1

Final velocity = V_i = 0 ms^-1

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

S₃ = V_av x t

S₃ = (V_f + V_i)/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

 

 

 

 

 

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 = V_f = 0 ms^-1

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

Since,      V_f = V_i + gxt

V_i = V_f – gxt

V_i = 0 – (-10) x 3

                              V_i = 30 ms^-1

Now using 3^rd equation of motion

2aS = V_f2 – V_i2

S =       V_f2 – V_i2/2a

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 = V_{i =} 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 = V_f² – V_i²

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:

V_f = 0              ,           S₂ =?

2aS = V_f² – V_i²

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

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

S₁ = 1600/6

S₂ = 266.66m

 

 

 

 

 

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 = V_i = 96 kmh^-1 = 96 x 1000/3600 = 96000/3600 ms^-1

Final velocity = V_f = 0 ms^-1

a = -(48000/3600)² x 3/1600 ms^-2

time = t =?

Or   V_f = V_i + at

Or   at = V_f – V_i

t = V_f – V_i/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