Question

A skier starts from rest and slides 9.0m down a slope in 3.0s. In what time after starting will the skier acquire a speed of 24m/s? Assume that the acceleration is constant.

Answer 1

The distance $h = 9m$ covered by the skier under a constant acceleration from the rest is given by the kinematic equation:

$$h =\dfrac{at_0^2}{2}$$

where $a$ is the acceleration and $t_0 = 3s$ is the time of covering $h$. Thus, obtain:

$$a = \dfrac{2h}{t_0^2}$$

On the other hand, by definition, the acceleration is:

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

where $v_f = 24m/s,v_i = 0m/s$ are the final and initial speeds respectively, and $t$ is the time when the skier acquire a speed $v_f$. Expressing $t$ and substituting $a$ obtain:

$$t = \dfrac{v_f}{a} = \dfrac{v_ft_0^2}{2h}\\ t = \dfrac{24\cdot 3^2}{2\cdot 9} = 12s$$

Answer. 12s.