Question

A train that is 268.7 m long undergoes constant acceleration the moment the last car is outside of the station, how far is the front of the train from the station after 25.0 s if it’s initial speed before acceleration is 4.48 m/s and it’s final speed is 27.4 meters per second ?

Answer 1

At the initial time when t=0 the front of the train is at a distance of 268.7 m from the station and starts at the speed of 4.48 m/s.

Now we find acceleration by applying the formula a= change in v/t.

Therefore, we get a= 0.917m/s2.

Now to get the position we use the formula x=x0+ final speed whole square – initial speed whole square / 2a.

We get the answer as 667m.