A train is traveling south at 23.5 m/s when the brakes are applied. It slows down with a constant rate to a speed of 5.01 m/s in a time of 8.73 s. How far does the train travel during the 8.73 s?

Question

I have already found
-Acceleration during braking: -2.12m/s

Answer 1

vf^2=vi^2+2ad
put in your vf, vi, and a solve for d.

Answer 2

Thank you! I was stuck on this problem

Answer 3

So,

vf = final velocity = 5.01m/s
vi = initial velocity = 23.5m/s
a = acceleration = -2.12m/s^2
d = distance

I plugged everything in and got 0.045.
Is this the correct way to solve?

5.01^2 = 23.5^2 + 2(-2.12)d
25.1001 = 548.01d
d = 25.1001/548.01

Thanks!

Answer 4

I didn't check your math, but that is the way, you are not paying attenion to sig figures.