Question

A 108 m long train starts from rest (at t = 0) and accelerates uniformly. At the same time (at t = 0), a car moving with constant speed in the same direction reaches the back end of the train. At t = 12 s the car reaches the front of the train. However, the train continues to speed up and pulls ahead of the car. At t = 32 s, the car is left behind the train. Determine, the car’s speed the train’s acceleration

Answer 1

Speed of the car = V

Length of the train = L = 108 m

Initial speed of the train = V₁ = 0 m/s

Acceleration of the train = a

T₁ = 12 sec

Distance traveled by the train in 12 sec = D₁

D₁ = V₁T₁ + aT₁²/2

D₁ = (0)T₁ + aT₁²/2

D₁ = aT₁²/2

Distance traveled by the car in 12 sec = D₂

D₂ = VT₁

The car starts at back end of the train and reaches the front end at 12 sec.

D₂ = D₁ + L

VT₁ = aT₁²/2 + L

V(12) = a(12)²/2 + 108

12V = 72a + 108

V = 6a + 9

At time t=32 sec the car is left behind the train.

T₂ = 32 sec

Distance traveled by the train in 32 sec = D₃

D₃ = V₁T₂ + aT₂²/2

D₃ = (0)T₂ + aT₂²/2

D₃ = aT₂²/2

Distance traveled by the car in 32 sec = D₄

D₄ = VT₂

The distance traveled by the train and the car in 32 sec is same as the car is at the back end of the again in 32 sec.

D₃ = D₄

VT₂ = aT₂²/2

V = aT₂/2

6a + 9 = a(32)/2

6a + 9 = 16a

10a = 9

a = 0.9 m/s²

V = 6a + 9

V = (6)(0.9) + 9

V = 14.4 m/s

a) Speed of the car = 14.4 m/s

b) Acceleration of the train = 0.9 m/s²