Question

As a train accelerates away from a station, it reaches a speed of 4.7 m/s in 5.0 s. If the train's acceleration is constant, what is its speed after an additional 6.0 s have elapsed?

Answer 1

Step-by-step explanation:

Acceleration = change in velocity / change in time

= 4.7 m/s / 5 s = .94 m/s^2

v = a t = (.94 m/s^2) ( 4.7 + 6.0) = ~ 10.1 m/s