Question

1. A subway train starts from rest at a station and accelerates at a rate of 2.50 m/s ^ 2 for 15.90 seconds. It then runs at a constant speed for 35 seconds and then slows down at a rate of 4.50 m/s ^ 2 until it stops at the next station .

(a) What is the speed of the train just before it starts slowing down ? (Ans .: v=39.75 m/s)

(b) How long did it take the train to finish the trip from one station to the other ? (Ans .: t=59.73 s)

Answer 1

1.a)

Let us consider the motion of the train for the first 15.90 seconds.

We have:

Initial velocity because it starts from rest

Acceleration

Time

We have the equation of motion for the final velocity given by:

Substituting the values:

This is the speed of the time when it stops accelerating. Since it then runs at a constant speed until it starts slowing down, the speed of the train just before it starts slowing down is equal to it. Therefore,

b)

Let us now consider the motion of the train from when it starts slowing down until it stops at the next station.

We have:

Initial velocity which is the velocity when it starts slowing down

Final velocity because it comes to rest at the station

Acceleration where the minus sign indicates that it is slowing down.

Let us again use the equation

Rearranging for the time,

Substituting,

This is the time for which it slows down

The total time taken to finish the trip is the sum of the time spent accelerating, time it runs at constant speed and the time it slows down.

Therefore,

The total trip time is 59.73s