Question

A man increased his speed by 5m/s after 30s,if his initial velocity was 20m/s.the average distance covered if his speed was at uniform acceleration would be?

Answer 1

The distance covered by uniform acceleration is:

$$d = v_0t + \dfrac{at^2}{2}$$

where $v_0 = 20m/s$ is the initial speed, $t = 30s$ is the time of motion, $a$ is the uniform acceleration.

By definition, the acceleration is:

$$a = \dfrac{\Delta v}{t}$$

where $\Delta v = 5m/s$ is the increase in the speed in time $t$. Thus, obtain:

$$d = v_0t + \dfrac{\Delta vt^2}{2t} = v_0t + \dfrac{\Delta vt}{2} = \left(v_0 + \dfrac{\Delta v}{2} \right)t$$

In numbers:

$$d = \left(v_0 + \dfrac{\Delta v}{2} \right)t = \left(20 + \dfrac{5}{2} \right)\cdot 30 = 675m$$

Answer. 675 m.