Question

A cheetah moving with constant acceleration covers the distance between two points 85.5 m apart in 9.5 s. It s speed as it passes the second point is 16 m/s.What is it s speed at the first point? what is its acceleration?

Answer 1

The distance traveled by a cheetah

$$d=\frac{v_i+v_f}{2}t$$

So

$$85.5=\frac{v_i+16}{2}\times 9.5$$

The initial velocity

$$v_i=2\:\rm{m/s}$$

The acceleration

$$a=\frac{v_f-v_i}{t}=\frac{16-2}{9.5}=1.47\:\rm{m/s^2}$$