Question

A tram leaves a station and accelerates for 2 minutes until it reaches a speed of 12 metres per

second. It continues at this speed for 1 minute.

It then decelerates for 3 minutes until it stops at the next station.


Calculate the distance, in metres, between the two stations.​

Answer 1

Explanation:

Acceleration with which the train moves:

v = u + at

12 = 0 + a(120)

a = 0.1m/s^2

Distance covered by the train in first part:

S = 1/2at^2

S = 1/2×0.1×(120)^2

S = 720m

Distance covered in second part:

S = ut

S = 12×60 = 720m

Deceleration at which train moves:

v = u+ at

0 = 12+a(180)

a = -1/15

Distance covered by the train in last part:

S = ut- 1/2at^2

S = 12(180) - 1/2(1/15)(180)^2

S = 2160 - 1080

S = 1080m

Therefore, the total distance between the two stations is = 720+720+1080 = 2520m.