Question

On turning a corner a car driving at $36\,km{h^{ - 1}}$ finds a child on the road $55\,m$ ahead. He immediately applies brakes, so as to stop within $5\,m$ of the child. Calculate the retardation produced and the time taken by the car to stop.

Answer 1

Hint:To solve this question first we convert the unit of speeds to our desired unit. Then by reading the question we will find the distance traveled by car before stopping. so, now we know the car's initial speed and final speed as well as the distance covered by the car. We will use the equation of kinematics and calculate the retardation. After calculating it we can calculate the time taken to reach the final speed from the initial speed.

Complete step by step answer:
Given, Car running at speed $ = 36\,km{h^{ - 1}}$
Child runs on the road at 55m ahead.
Cars stop within 5m of the child.
Car travel distance before stopping $ = 55 - 5 = 50\,m$
Initial speed $ = 36\,km{h^{ - 1}} = \dfrac{5}{{18}} \times 36 = 10\,m{s^{ - 1}}$

We have to calculate the acceleration:
So, use:
${v^2} - {u^2} = 2as$
Keeping value in it,
$ \Rightarrow {0^2} - {(10)^2} = 2 \times 50 \times a$
Further solving it:
$ \Rightarrow - 100 = 100 \times a$

By cross multiplication:
$ \Rightarrow a = - \dfrac{{100}}{{100}} = - 1\,m{s^{ - 2}}$
So it is minus mean retardation.
To calculate time taken:
Use, $v = u + at$
keeping value in it. We get,
$0 = 10 + ( - 1)t$
By cross multiplication:
$ \therefore 10\sec = t$

Therefore, time taken will be 10 sec.

Note:We can solve any kinematic problem by using three equations of motion. We just have to first find what we have to calculate and what is given in the question, then choose the right equation and calculate it. so, always try to simplify the given data in question as well as we should keep in mind the units. as we can operate the same units only.When we have calculated time then we can also use the second equation of kinematics but then it will be a little bit harder. So, we used the first one.