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

Answer:

66.28 s

Explanation:

Step 1:

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

Time,t=9.75 min=$9.75\times 60=585s$

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

$0-41.425=-0.625t$

$-41.425=-0.625t$

$t=\frac{-41.425}{-0.625}$

$t=66.28s$

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