Kinematics – 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
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.
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?
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.
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
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.
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:
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
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)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 = 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
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