Answer to Question #149055 in Physics for Melanie Belandres

Question

3. A car moving with constant acceleration covers the distance between two points 60 m apart in 6 s. Its velocity as it passes the second point is 15 m/s.
a) What is its velocity at the first point?
b) What is the acceleration?

Answer 1

$s=\frac{v^2-v_0^2}{2a}$ and $v=v_0+at$ $\to$ $v_0=v-at$

$s=\frac{v^2-(v_0-at)^2}{2a}\to a=\frac{2vt-2s}{t^2}=\frac{2\cdot 15\cdot 6-2\cdot60}{6^2}=1.67(m/s^2)$

$v_0=v-at=15-1.67\cdot6=5(m/s)$

Answer. $v_0=5(m/s); a=1.67(m/s^2)$