Question

If the train can slow down at a rate of 0.625 m/s2, how long, in seconds, does it take to come to a stop from this velocity?

Answer 1

66.28 s

Step-by-step explanation:

Step 1:

Acceleration,a=
`0.065m/s^2`

Time,t=9.75 min=
`9.75* 60=585s`

1 min=60 s

Initial velocity,u=
`3.4m/s`

We have to find the final velocity of train.

We know that

`v=u+at`

Substitute the values

`v=3.4+0.065(585)=41.425m/s`

Step 2:

Now, initial velocity,u=41.425m/s

Deceleration,a=
`-0.625m/s^2`

Because the train velocity decreases

Final velocity, v=0

Again, substitute the values in the above formula

`0-41.425=-0.625t`

`-41.425=-0.625t`

`t=(-41.425)/(-0.625)`

`t=66.28s`

Hence, the train takes 66.28 s to come to stop from velocity 41.425m/s.