Question

The minimum stopping distance for a car traveling at a speed

of 30 m/s is 60 m, including the distance traveled during the

driver’s reaction time of 0.50 s. What is the minimum stopping distance for the same car traveling at a speed of 40 m/s?

Short answer

Minimum stopping distance travelled by the car at a speed of$40m/s$is$100m.$

Step-by-step solution

Given information

Speed of the car is $u=30m/s.$

Stopping distance of the car with this speed is$D=60m.$

Reaction time taken by the driver, $t=0.50s.$

Calculation of the distance travelled while deacceleration

During reaction time, the distance travelled by the driver is,

$$\begin{gathered} x=ut \\ x=\left(30\times 0.50\right)m \\ x=15m \end{gathered}$$

Therefore, while slowing down, distance travelled by the driver is,

$$\begin{gathered} s=D-x \\ s=\left(60-15\right)m \\ s=45m \end{gathered}$$

Calculation of stopping distance

Initial acceleration of the car can be calculated as,

$$\begin{gathered} 0=u^2+2as \\ a=-\frac{u^2}{2s} \\ a=-\frac{{30}^2}{2\times 45} \\ a=-10m/s^2 \end{gathered}$$

Now, considering the initial speed as $u\text{'}=40m/s,$distance travelled by the car during constant motion is,

$$\begin{gathered} s_1=u\text{'}t \\ {s}_1=40\times 0.5m \\ {s}_1=20m \end{gathered}$$

The distance travelled by the car during slowing down can be obtained as,

$$\begin{gathered} 0=u{\text{'}}^2+2a{s}_2 \\ {s}_2=-\frac{u{\text{'}}^2}{2a} \\ {s}_2=-\frac{{40}^2}{2\left(-10\right)}m \\ {s}_2=80m \end{gathered}$$

Therefore, minimum stopping distance should be,

$$\begin{gathered} S=s_1+s_2 \\ S=\left(20+80\right)m \\ S=100m \end{gathered}$$