Question: An automobile driver increases the speed at a constant rate from 25 km.hrto 55 km/hrin 0.50 min. A bicycle rider speeds up at a constant rate from rest to 30 km/hr in 0.50 min. What are the magnitudes of (a) the driver’s acceleration and (b) the rider’s acceleration?

Initial speed of driver, $v_0=25\text{km/h}$

Final speed of driver, $v=55\text{km/h}$

Initial speed of rider, ${v_0}^1=0\text{km/h}$

Final speed of rider, $v^1=30\text{km/h}$

Time interval, $t=0.50\text{min}$

The expression for the kinematic equations of motion are given as follows:

$v=v_0+at$ … (i)

Here, $v_0$is the initial velocity, $v$is the final velocity, $a$is the acceleration and$t$is the time.

Convert the velocity from km/h to m/s as:

$$\begin{gathered} v_0=25\times \frac{\text{km}}{\text{h}}\times \frac{1000\text{m}}{1\text{km}}\times \frac{1\text{h}}{3600\text{s}} \\ =6.94\text{m/s} \\ v=55\times \frac{\text{km}}{\text{h}}\times \frac{1000\text{m}}{1\text{km}}\times \frac{\text{1h}}{3600\text{s}} \\ =15.28\text{m/s} \end{gathered}$$

Using equation (i), the acceleration can be calculated as follows:

$$\begin{gathered} a=\frac{v-v_0}{t} \\ =\frac{15.28\text{m/s-6.94m/s}}{0.5\times 60\text{s}} \\ =0.28{\text{m/s}}^2 \end{gathered}$$

Thus, the acceleration of the driver is$0.28{\text{m/s}}^2$ .

Convert the velocity from km/h to m/s as:

$$\begin{gathered} v^1=30\times \frac{\text{km}}{\text{h}}\times \frac{1000\text{m}}{1\text{km}}\times \frac{1\text{h}}{3600\text{s}} \\ =8.33\text{m/s} \end{gathered}$$

Using equation (i), the acceleration can be calculated as follows:

$$\begin{gathered} a^1=\frac{v^1-{v_0}^1}{t} \\ =\frac{8.33\text{m/s-0m/s}}{0.5\times 60\text{s}} \\ =0.28{\text{m/s}}^2 \end{gathered}$$

Thus, the acceleration of the rider is$0.28{\text{m/s}}^2$ .