Question

A train takes 11 minutes to travel between two stations which are 9.6 km apart, starting from rest at
the one station and coming to rest at the other station. The train accelerates uniformly for the first
2 minutes and retards uniformly for the final 4 minutes, with the speed being constant for the rest
of the time.
1.1 Draw a neat, fully labelled, velocity vs. time graph of the journey between the two stations.
(5)
1.2 Determine the constant speed of the train.
(5)
1.3 Determine the acceleration of the train.
(3)
1.4 Determine the deceleration of the train.
(3)
1.5 Determine the distance that the train travelled while accelerating.
(1)
1.6 Determine the distance that the train travelled while decelerating.
(1)
1.7 Determine the relationship between the acceleration and the deceleration of the train.

Answer 1

Given : A train takes 11 minutes to travel between two stations which are 9.6 km apart, starting from rest at

the one station and coming to rest at the other station. The train accelerates uniformly for the first

2 minutes and retards uniformly for the final 4 minutes, with the speed being constant for the rest of the time.

To Find : Determine the constant speed of the train.  

Determine the acceleration of the train.  

Determine the deceleration of the train.

Determine the distance that the train travelled while accelerating.  

Determine the distance that the train travelled while decelerating.  

Determine the relationship between the acceleration and the deceleration of the train.

Solution:

9.6 km = 9600  m

let say acceleration is a

Then Sped achieved

V = 2a

Distance covered in 2 mins  = (1/2)a2² = 2a  m

Distance covered in next 5 minutes = 2a * 5 = 10a m

Distance covered in last 4 mins  =  4 * 2a/2  = 4a m

deceleration  in last 4 mins = ( 2a)/4 = a/2  m/s²

2a + 10a + 4a  = 16a  = 9600

=> a = 600 m/s²

constant speed of the train.  =2a  = 1200 m/s

acceleration of the train. = 600 m/s²

deceleration of the train. = 300 m/s²

distance that the train travelled while decelerating. = 4 * 6 = 2400 m

= 2.4 km

deceleration of the train. is half of the acceleration

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