Question

A car is travelling at $20\,m{s^{ - 1}}$ along a road. A child runs into out on the road $50\,m$ ahead and the car driver steps on the pedal. What must be the car’s deceleration if the car is to stop just before it reaches the child?

Answer 1

In this question, the acceleration would be negative since it is a retarding force. Also, we will take the distance covered by the car before stopping to be equal to the distance of the car from the child which is given to be $50\,m$ . Since we are to stop a moving car, the final velocity will be zero. Then we will suitably choose the speed equation and substitute the values in it to get the final answer.

Formula used:
The speed equations are
$v = u + at$
$\Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
$\Rightarrow 2as = {v^2} - {u^2}$
Where $u$ is the initial velocity, $v$ is the final velocity, $s$ is the distance covered, $t$ is the time taken and $a$ is the acceleration.

Complete step by step answer:
Since we are to stop a moving car, the final velocity will be zero.
Hence, $v = 0\,m{s^{ - 1}}$
Given that the initial velocity of the car is $u = 20\,m{s^{ - 1}}$
The distance covered before stopping is $s = 50\,m$ .
We will use the third speed equation of motion given as $2as = {v^2} - {u^2}$
Substituting the values in the equation,
$2a(50) = {0^2} - {20^2}$
Further solving this equation, we get,
$100a = - 400$
$\therefore a = - 4\,m\,{s^{ - 2}}$
Since acceleration is coming out to be negative, it is acting in a direction opposite to the direction of motion.

So, the car should retard at $a = - 4\,m\,{s^{ - 2}}$ to stop the car just before reaching the child.

Note: We must carefully notice the sign of the physical quantities given in the question. Here, the acceleration is negative which means that the direction of acceleration is opposite to the direction of motion. Hence, a sense of pull back effect is created when we apply brakes. The negative sign should not be dropped while doing calculations since it will lead to wrong answers.