Question: A cyclist is traveling at \[15m{s^{ - 1}}\]. She applied breaks so that she does not collide with a ...

Question

A cyclist is traveling at $15m{s^{ - 1}}$ . She applied breaks so that she does not collide with a wall $18m$ away. What deceleration must she have?

Answer 1

Solution

Deceleration is the opposite of acceleration. Deceleration is the decreased speed of a moving body. If the initial velocity, final velocity and the traveled distance is given, the deceleration of a moving body can be calculated from the equation of motion. In this problem, the final velocity will be zero to avoid collision with the wall.
Formula used:
Formula of deceleration $a = \dfrac{{{v^2} - {u^2}}}{{2s}}$
$u$ is the initial velocity, $v$ is the final velocity and, $s$ is the distance.

Complete answer:
The equation of motion of the car in terms of the velocities, acceleration and, displacement is
$a = \dfrac{{{v^2} - {u^2}}}{{2s}}$
Where, $a$ is the declaration, $u$ is the initial velocity, $v$ is the final velocity and, $s$ is the distance.
The final velocity is zero because the car has to avoid collision with the wall.
Given, the initial velocity of the body is $u = 15$ , the final velocity $v = 0$ , traveled distance $s = 18$ .
$\therefore a = \dfrac{{{v^2} - {u^2}}}{{2s}}$
$\Rightarrow a = \dfrac{{0 - {{15}^2}}}{{2 \times 18}}\;$
$a = - 6.25m/{s^2}$
She must have $6.25m/{s^2}$ deceleration.

Note:
We get the negative value. The sign implies that this is the declaration or retardation which is opposite to the acceleration. When a car has to be stopped the speed has to be reduced and finally zero. The decrease of speed at some interval defines the declaration. Whereas, acceleration is the rate of increase in velocity. The deceleration can be calculated by the negative value of acceleration.
From the equation of motion,
$\therefore a = \dfrac{{{v^2} - {u^2}}}{{2s}}$
$\Rightarrow {v^2} - {u^2} = 2as$
If the final velocity is zero and, for the constant declaration,
${u^2} \propto s$
So, the distance is directly proportional to the square of the velocity. So, for example, if the speed is doubled the car is tried to be stopped from four times the distance.