91Ó°ÊÓ

The average acceleration of the object in a certain time period is calculated by taking the ratio of the change in velocity to the total time taken, i.e.,

$\begin{array}{c}\mathrm{Average}\mathrm{acceleration}=\frac{\mathrm{Total}\mathrm{change}\mathrm{in}\mathrm{velocity}}{\mathrm{Total}\mathrm{time}\mathrm{taken}}\\ \stackrel{¯}{a}=\frac{\mathrm{Δ}v}{\mathrm{Δ}t}\end{array}$.

The initial velocity of the bicyclist at the crest of the mountain is

$\begin{array}{c}u=15\mathrm{km}/\mathrm{h}\\ =\left(15\mathrm{km}/\mathrm{h}\right)×\left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right)×\left(\frac{1\mathrm{h}}{3600s}\right)\\ =4.17\mathrm{m}/\mathrm{s}\end{array}$

The final velocity of the bicyclist at the bottom is

$\begin{array}{c}v=65\mathrm{km}/\mathrm{h}\\ =\left(65\mathrm{km}/\mathrm{h}\right)×\left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right)×\left(\frac{1\mathrm{h}}{3600s}\right)\\ =18.06\mathrm{m}/\mathrm{s}\end{array}$

The distance traveled by the bicyclist while riding down the mountain is

$\begin{array}{c}\mathrm{Δ}x=4.0\mathrm{km}\\ =\left(4.0\mathrm{km}\right)×\left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right)\\ =4000\mathrm{m}\end{array}$

Let $\mathrm{Δ}t$be the time taken by the bicyclist to reach the bottom of the mountain.

The average velocity of the bicyclist during the motion is

$\begin{array}{c}\stackrel{¯}{v}=\frac{u+v}{2}\\ =\frac{\left(4.17+18.06\right)\mathrm{m}/\mathrm{s}}{2}\\ =11.11\mathrm{m}/\mathrm{s}\end{array}$

The average velocity can be written as

$\stackrel{¯}{v}=\frac{\mathrm{Δ}x}{\mathrm{Δ}t}$

Thus, the time taken by the bicyclist to reach the bottom is

$\begin{array}{c}\mathrm{Δ}t=\frac{\mathrm{Δ}x}{\stackrel{¯}{v}}\\ =\frac{4000\mathrm{m}}{11.11\mathrm{m}/\mathrm{s}}\\ =360.04\mathrm{s}\end{array}$

The average acceleration of the bicyclist while riding down the mountain is

$\begin{array}{c}\stackrel{¯}{a}=\frac{\mathrm{Δ}v}{\mathrm{Δ}t}\\ =\frac{v-u}{\mathrm{Δ}t}\\ =\frac{\left(18.06-4.17\right)\mathrm{m}/\mathrm{s}}{360.04\mathrm{s}}\\ =0.039\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, the average acceleration of the bicyclist while riding down the mountain is $0.039\mathrm{m}/{\mathrm{s}}^2$.