Question

A car moving with constant acceleration covers the distance between two points 60 m apart in 6 sec. It speed as its pass the second point is 15 m/s. what is its speed at the first point?

Answer 1

By definition the acceleration is given as follows:

$$a = \dfrac{v_f - v_i}{t}$$

where $v_f = 15m/s$ is the final speed, $v_i$ is the initial speed, and $t = 6s$ is the time.

The distance travelled under the constant acceleration is given as follows:

$$d = v_it + \dfrac{at^2}{2}$$

Substituting the expression for $a$, have:

$$d = v_it + \dfrac{(v_f - v_i)t}{2} = \dfrac{(v_i+v_f)t}{2}\\$$

Expressing $v_i$, get:

$$v_i = \dfrac{2d}{t} - v_f = \dfrac{2\cdot 60m}{6s} - 15m/s = 5m/s$$

Answer. 5 m/s.