Question

A driver driving a truck at a constant speed of 20m/s suddenly saw a parked car ahead of him at a distance of 95m. He applies the brakes after sometime to produce retardation of \[2.5m/{s^2}\]. If an accident was avoided, his reaction time is​?
A. 0.5 s
B. 0.75 s
C. 0.8 s
D. 1 s

Answer 1

Hint: As some of the quantities of the motion are given, and time has to be calculated, we can use Newton's equations of motion to find the required value. The reaction time will be based upon the distance left to be covered when the body comes to rest after a specific period of time.

Formula used:
1) \[v = u + at\] where, v and u are final and initial velocities respectively, a is acceleration and t is time taken.
2) $s = ut + \dfrac{1}{2}a{t^2}$ where s is the distance covered.

Complete step by step answer:
It is given that the driver was moving initially at a constant speed of 20m/s, then applies the brake to finally stop the vehicle (vehicle comes at rest). Retardation (negative acceleration) produced is of magnitude \[2.5m/{s^2}\].
From Newton’s first equation of motion:
\[v = u + at\] here,
Initial velocity (u) = 20 (given)
Final velocity (v) = 0 (come to rest)
Acceleration (a) = - 2.5 (negative because it is retarded)
Time (t) = t (say)
\[
0 = 20 - 2.5t \\
\Rightarrow 2.5t = 20 \\
\Rightarrow t = \dfrac{{20}}{{2.5}} \\
\Rightarrow t = \dfrac{{20 \times 10}}{{25}} \\
\Rightarrow t = 8s \\
\]
Thus the truck took 8 seconds to come to rest.
Now, calculating the distance driver can cover within 8 seconds using Newton’s second equation of motion:
$s = ut + \dfrac{1}{2}a{t^2}$
Substituting the known values to find the distance s:
$
s = \left( {20 \times 8} \right) + \dfrac{1}{2} \times \left( { - 2.5} \right) \times {\left( 8 \right)^2} \\
\Rightarrow s = \left( {20 \times 8} \right) - \dfrac{1}{2} \times 2.5 \times 64 \\
\Rightarrow s = 160 - 80 \\
\Rightarrow s = 80m \\
$
The truck can cover 80 metres in 8 seconds, but the required distance to be covered is 95 metres (given). So the extra distance is:
$95 - 80 = 15m$
Thus, the driver’s reaction time will be based on this distance and his initial speed 20 m/s:
\[
S = \dfrac{D}{T} \\
\Rightarrow T = \dfrac{D}{S} \\
\Rightarrow T = \dfrac{{15}}{{20}} \\
\therefore T = 0.75s \\
\]
Therefore, the reaction time of the truck driver is 0.75 seconds and the correct option is B.

Note:We decide which equation of motion is to be used by observing the quantities that are given and which needs to be calculated and choose the equation which satisfies the needs. We used the remaining distance to calculate reaction time because it is given that the accident was just avoided but the distance the truck was covering before coming to rest was lesser compared to the given distance. So, the distance of his reaction time would be the difference between the two.