Question

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?

Short answer

a) The driver’s acceleration is$0.28{\text{m/s}}^2$ .

(b) The rider’s acceleration is $0.28{\text{m/s}}^2$.

Step-by-step solution

Given Data

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, $t=0.50\text{min}$

Understanding the kinematic equations Kinematic equations describe the motion of an object with constant acceleration. These equations can be used to determine the acceleration. 

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.

: (a) Determination of the acceleration of the driver 

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$0.28{\text{m/s}}^2$ .

(b) Determination of the acceleration of the rider

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$0.28{\text{m/s}}^2$ .