Answer to Question #171114 in Mechanics | Relativity for Ria Bandillon Gaddi

Question

a train changes its speed uniformly from 60 mph to 30 mph in a distance of 1,500 ft. what is its acceleration

Answer 1

For the motion with constant acceleration the distance is

$s(t) = v_0t + \dfrac{at^2}{2}.$

The velocity may be written in form $v(t) = v_0 + at,$ so $t = \dfrac{v(t)-v_0}{a}$ .

Therefore, $s = \dfrac{v_0(v(t)-v_0)}{a} + \dfrac{(v(t)-v_0)^2}{2a} = \dfrac{v(t)^2-v_0^2}{2a}$

So $a= \dfrac{v(t)^2-v_0^2}{2s} = \dfrac{(30\cdot1.4667)^2-(60\cdot1.4667)^2 }{2\cdot1500} = -1.94\,\mathrm{ft/s^2}$