Question

A reaction time for an automobile driver is \[0.7\] seconds. If the automobile can be decelerated at \[0.5\,{\text{m}}{{\text{s}}^{ - 2}}\] . Calculate the total distance travelled in coming to stop from an initial velocity of \[8.33\,{\text{m}}{{\text{s}}^{ - 1}}\] after signal is observed.
A. \[12.77\,{\text{m}}\]
B. \[14.82\,{\text{m}}\]
C. \[16.83\,{\text{m}}\]
D. \[19.65\,{\text{m}}\]

Answer 1

Hint: First of all, we calculate the distance travelled by the car in \[0.7\] seconds. Then we will again calculate the distance moved by the car with the deceleration given in the question. At last, we will add the two distances to find the correct answer

Complete step by step answer:
In the given question, we are supplied with the following data:
The time for which the driver takes his time to apply the brakes is \[0.7\] seconds.
The deceleration of the car was \[0.5\,{\text{m}}{{\text{s}}^{ - 2}}\] .
The initial velocity of the car was given as \[8.33\,{\text{m}}{{\text{s}}^{ - 1}}\] .
We are asked to calculate the distance for which the car moves after the brakes have been applied before coming to complete halt.
To begin with, let us first understand what the situation is.
The automobile must come to a complete halt. So, the acceleration here in this case will be negative, as the car gradually decreases its speed to make the stop. But the car does not suddenly make the stop after the brakes have been applied. It therefore moves to a certain distance.
First, we will calculate the distance moved by the car before applying the brakes, which can be found out by the formula given below:
\[S' = u \times t\] …… (1)
Where,
\[S'\] indicates the distance moved by the car before the driver reacts.
\[u\] indicates initial velocity of the car.
\[t\] indicates the reaction time of the driver.
We will substitute the requisite values in the equation (1) and we get:
$S' = u \times t \\
\implies S' = 8.33 \times 0.7\,{\text{m}} \\
\implies S' = 5.83\,{\text{m}} \\$
Now, we will find out the distance moved by the car after the brakes have been applied. We will use one of the formulae from laws of motion, which is given below:
\[{v^2} = {u^2} + 2as\] …… (2)
Where,
\[S\] indicates the distance moved by the car after the brakes have been applied, before coming to halt.
\[v\] indicates the final velocity which is zero in this case.
\[u\] indicates initial velocity of the car.
\[a\] indicates deceleration of the car.
Now, we substitute the required values in the equation (2), and we get:
${v^2} = {u^2} + 2as \\
\implies {0^2} = {\left( {8.33} \right)^2} + 2 \times \left( { - 5} \right) \times S \\
\implies S = \dfrac{{69.4}}{{10}} \\
\implies S = 6.94\,{\text{m}} \\$
Therefore, the total distance moved by the car is:
$S + S' \\
= \left( {5.83 + 6.94} \right)\,{\text{m}} \\
= 12.77\,{\text{m}} \\$
Hence, the total distance travelled by the car is \[12.77\,{\text{m}}\] .

So, the correct answer is “Option A”.

Note:
In this given problem, many students seem to make mistakes while finding the distance. They simply take the time as \[0.7\] seconds and put in the formula and get a wrong answer. First you have to calculate the distance moved by the car when the driver takes \[0.7\] seconds to react before applying brakes.