Question

A cyclist travelling at $15\,m{s^{ - 1}}$ applies brakes to avoid collision with a car $18\,m$ away. What deceleration must he have?
(A) $8.25\,m{s^{ - 2}}$
(B) $82.5\,m{s^{ - 2}}$
(C) $6.25\,m{s^{ - 2}}$
(D) $62.5\,m{s^{ - 2}}$

Answer 1

Hint: The deceleration of the cycle is given by using the equation of the motion, the given velocity is taken as the initial velocity and the final velocity is assumed to be zero, because the cycle finally stops, then only the deceleration can be determined.

Useful formula
The equation of the motion is given by,
${v^2} - {u^2} = 2as$
Where, $v$ is the final velocity of the cycle, $u$ is the initial velocity of the cycle, $a$ is the acceleration or deceleration of the cycle and $s$ is the distance between the car and the cycle.

Complete step by step solution
Given that,
The initial velocity of the cycle is, $u = 15\,m{s^{ - 1}}$,
The distance of the car and the cycle is, $s = 18\,m$.
Now, the deceleration can be determined by,
The equation of the motion is given by,
${v^2} - {u^2} = 2as\,..................\left( 1 \right)$
By substituting the initial velocity of the cycle and final velocity of the cycle and the distance between the car and the cycle in the above equation (1), then the above equation (1) is written as,
$\Rightarrow {0^2} - {15^2} = 2 \times a \times 18$
By using the square in the terms in the above equation, then
$\Rightarrow - 225 = 2 \times a \times 18$
By multiplying the terms in the RHS, then the above equation is written as,
$\Rightarrow - 225 = a \times 36$
By keeping the acceleration in one side and the other terms in the other side, then the above equation is written as,
$a = \dfrac{{ - 225}}{{36}}$
By dividing the terms in the above equation, then the above equation is written as,
$\Rightarrow a = - 6.25\,m{s^{ - 2}}$
The negative sign indicates the deceleration.

Hence, the option (C) is the correct answer.

Note: The negative sign of the final answer indicates that the cycle is decelerating, so that the cycle will stop. If the deceleration of the cycle is equal to the final answer, then the cycle will stop without colliding with the car, if the deceleration is not equal to the final answer, then the cycle will collide the car.