Question

On turning a corner, a motorist rushing at $44\,m{s^{ - 1}}$ finds a child on the road $100\,m$ ahead. He instantly stops the engine and applies the brakes so as to stop it within $1\,m$ of the child. Calculate time required to stop it.

Answer 1

Hint The time required can be determined by using the equation of the motion, the acceleration equation of the motion can be used to find the deceleration of the vehicle and the deceleration can be used to determine the time taken to stop the vehicle can be determined.
Useful formula:
The equation of the motion can be given as,
${v^2} - {u^2} = 2as$
Where, $v$ is the final velocity of the vehicle, $u$ is the initial velocity of the vehicle, $a$ is the acceleration of the vehicle and $s$ is the distance.
The equation of the motion is given as,
$v = u + at$
Where, $v$ is the final velocity of the vehicle, $u$ is the initial velocity of the vehicle, $a$ is the acceleration of the vehicle and $t$ is the time taken.

Complete step by step answer
Given that,
The initial velocity of the vehicle is given as, $u = 44\,m{s^{ - 1}}$,
The total distance between the vehicle and the child is, $100\,m$,
The distance between the vehicle and the child when the vehicle stops, $1\,m$,
Hence, the distance is, $s = 100 - 1 = 99\,m$.
The final velocity is assumed to be zero because the vehicle stops.
Now,
The equation of the motion can be given as,
${v^2} - {u^2} = 2as$
By substituting the initial velocity, final velocity and the distance in the above equation, then the above equation is written as,
${0^2} - {44^2} = 2a \times 99$
By squaring the terms in the above equation, then the above equation is written as,
$ - 1936 = 2a \times 99$
By multiplying the terms in the above equation, then the above equation is written as,
$ - 1936 = 198a$
By rearranging the terms in the above equation, then the above equation is written as,
$a = \dfrac{{ - 1936}}{{198}}$
By dividing the terms in the above equation, then the above equation is written as,
$a = - 9.7\,m{s^{ - 1}}$
Now,
The equation of the motion is given as,
$v = u + at$
By substituting the initial velocity, final velocity and the acceleration in the above equation, then the above equation is written as,
$0 = 44 + \left( { - 9.7} \right)t$
By rearranging the terms in the above equation, then the above equation is written as,
$9.7t = 44$
By rearranging the terms in the above equation, then the above equation is written as,
$t = \dfrac{{44}}{{9.7}}$
By dividing the terms in the above equation, then the above equation is written as,
$t = 4.5\,\sec $

Note The time taken by the object is directly proportional to the velocity of the object and the time taken by the object is inversely proportional to the acceleration of the object. As the velocity of the object increases, then the time taken by the object is also increasing.