Table of Contents
Numerical Problems
2.1 A train moves with a uniform velocity of 36 kmh1 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 t^{2}
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 t^{2}
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 = S_{1} + S_{2 }
By using this relation
S_{1} = V_{t}
S_{1} = 40 x 5
S_{1} = 200 m ………………………………. (i)
Now by using 3^{rd} equation of motion
2aS_{2} = V_{f}^{2} – V_{i}^{2}
S_{2} = V_{f}^{2} – V_{i}^{2}/2a
S_{2} = (0)2 – (40)2/2 x (4)
S_{2} = 1600/8
S_{2} = 200 m ……………………………………… (ii)
From (i) and (ii) we get;
S = S_{1} + S_{2}
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 kmh1, 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 = Vf^{2} – V_{i}^{2}
^{ }2 x 0.5 x 100 = Vf^{2} – 0
Or , 100 = Vf^{2}
^{ }Or Vf^{2 }= 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}
S_{1} = V_{av} x t
S_{1} = (V_{f} + V_{i})/2 x t
S_{1} = 13.333 + 0/2 x 120
S_{1} = 6.6665 x 120
S_{1} = 799.99 m = 800 m
Case – II:
Uniform velocity = V_{f} = 13.333 ms^{1}
Time = t = 5 minutes = 5 x 60 = 300 s
S_{2} = v x t
S_{2} = 13.333 x 300
S_{2} = 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
S_{3} = V_{av} x t
S_{3} = (V_{f }+ V_{i})/2 x 180
S_{3} = 13.333 + 0/2 x 180
S_{3} = 6.6665 x 180
S_{3} = 1199.97 = 1200 m
Total distance = S = S_{1} + S_{2 }+ S_{3}
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 ms1)
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 kmh1 to 48 kmh1 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 = S_{1 }=?
First of all, we will find the value of acceleration a
2aS = V_{f}^{2} – V_{i}^{2}
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 S^{2}:
V_{f }= 0 , S_{2 }=?
2aS = V_{f}^{2} – V_{i}^{2}
2 (48000/36000)^{2} x 3/1600 x S_{1} = (0)^{2} – (48000/36000)^{2 }
S_{1} = (48000/36000)^{2} x (48000/36000)^{2} x 1600/3 x 2
S_{1} = 1600/6
S_{2} = 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
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