Question

The brakes applied to a car produce an acceleration of $6m/s^2$ in the opposite direction to the motion. If the car takes $2 s$ to stop after the application of brakes, calculate the distance it travels during this time.
A) 12 m
B) 10 m
C) 8 m
D) 18 m

Answer 1

When brakes are applied, the body will start retardation with uniform acceleration in opposite direction. Apply the equation of motion with proper sign convention.

Complete step by step solution:
After brakes are applied , the body will start slowing down and finally the body will stop which is due to retardation.
Retardation is the opposite of acceleration. It is the rate at which an object slows down. Retardation is the final velocity minus the initial velocity, with a negative sign in the result because the velocity is decreasing
GIven:
$\Rightarrow$ acceleration$(a) = - 6 m/s^2$
$\Rightarrow$ Time$(t) = 2s$
Final velocity, $v = 0m/s$
$\Rightarrow v = u + at$
$\Rightarrow 0 = u - 6 \times 2$
$\therefore u = 12 m/s$
$\Rightarrow s = ut + \dfrac{1}{2} \times at^2$
$\Rightarrow s = 12 \times 2 + \dfrac{1}{2}\times (-6) \times 4$
$\therefore s = 12m$
Retardation takes the negative value of acceleration as the motion will be in opposite direction

Thus option A is correct.

Additional information:
The formula for acceleration can be used to identify that the end result should contain a negative sign. And if time is also given, then retardation = (final velocity - initial velocity) / time.
Velocity is how fast an object moves. Thus, velocity is the change in the position of an object divided by time.
Velocity = (final position - initial position) / time
For one dimension motion we have three equations which can be used to find the final velocity , initial velocity, distance travelled, only if acceleration produced during motion is constant. Equations are as follows
$\Rightarrow {{1)}}{{{v}}^{{2}}}{{ = }}{{{u}}^{{2}}}{{ + 2as}}$
$\Rightarrow {{2)s = ut}}+\dfrac{{{1}}}{{{2}}}{{a}}{{{t}}^{{2}}}$
$\Rightarrow {3)v = u + at}$

Note: $s= \dfrac{u^2}{2a}$
This is an expression which can be directly used for finding the value of stopping distance
For a freely falling body, the equation of motion is applicable as the motion will be under the influence of constant acceleration due to gravity, which is equal to –g.