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:

I think so in this que, direct and inverse proportion formula will be applied -

u - 4.7m/s^2

t1 - 5sec

t2 - 5+6 = 11sec

4.7/5 = v/11

by cross multiplication we get - 4.7*11=5v

v = 10.34m/s