Question

A driver takes $0.20s$ to apply the brake after he sees a need for it. This is called the reaction time of the driver. If he is driving a car at a speed of $54km/h$and the brakes cause a deceleration of $6m/s^2$. Find the distance traveled by car after he sees the need to put the brakes on.

Answer 1

Step 1: Given

Reaction time: $t=0.20s$

Initial velocity of car: $u=54km/h$

Final velocity of car: $u=0m/s$ (as it comes to rest)

Acceleration: $a=-6m/s^2$ (since, it is deceleration)

Step 2: Formula Used

Distance is the product of speed and time, $dis\tan ce=speed\times time$.

According to the third equation of motion, $v^2=u^2+2as$, where $v$ is final velocity, $u$ is initial velocity, $a$ is acceleration and $s$ is displacement.

Step 3: Calculate the distance travelled by car in reaction time.

Calculate the initial velocity of car in meters per second by multiplying it by $\frac{5}{18}$.

$\begin{array}{rcl}u& =& 54km/h\\ & =& 54\times \frac{5}{18}m/s\\ & =& 15m/s\end{array}$

Calculate the distance travelled by car in reaction time.

$\begin{array}{rcl}dis\tan ce& =& speed\times time\\ & =& 15m/s\times 0.2s\\ & =& 3m\end{array}$.
Step 4: Calculate the total distance travelled by car

Calculate the distance travelled by car after the driver realizes the need to put the brakes on.

$\begin{array}{rcl}{v}^2& =& u^2+2as\\ & \Rightarrow & 0={15}^2+2\left(-6\right)s\\ & \Rightarrow & 0=225-12s\\ & \Rightarrow & 12s=225\\ & \Rightarrow & s=\frac{225}{12}\\ & \Rightarrow & s=18.75m\end{array}$.

Calculate the total distance travelled by car by adding both the distances.

$\begin{array}{rcl}Totaldis\tan ce& =& 3m+18.75m\\ & =& 21.75m\end{array}$

Hence, the total distance travelled by car is $\begin{array}{rcl}& & 21.75m\end{array}$.