Question

A train running along a straight track at 30m/s is slowex unformly to a stop in 44s. Find the acceleration and the stopping distance

Answer 1

The acceleration:

$$a=\frac{0-v}{t}=\frac{0-30}{44}=-0.68\frac{m}{s^2}$$

The stopping distance:

$$d=0.5vt=0.5(30)(44)=660\ m$$