A train is traveling at 44 m/s as it approaches a station. If the train's normal acceleration is 2.2 m/s2 and it takes 27 s for the train to come to a complete stop, how far in front of the station should the engineer apply the brakes?

s = 1989.9 m

Explanation:

In order to find the distance at which the engineer should apply brake, we can use 2nd equation of motion. The second equation of motion is as follows:

s = Vi t + (0.5)at²

where,

s = distance from the station = ?

Vi = Initial velocity = 44 m/s

t = time required to stop = 27 s

a = acceleration of train = 2.2 m/s²

Therefore,

s = (44 m/s)(27 s) + (0.5)(2.2 m/s²)(27 s)²

s = 1188 m + 801.9 m

s = 1989.9 m

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.