Question

A motor car running at the rate of $7m/s$ can be stopped by applying brakes in $10\text{ }m$. Show that total resistance to the motion, when brakes are on, is one fourth of the weight of the car.

Answer 1

With the acceleration not given, we can assume that the acceleration has been taken uniform. To find the resistance force of the brake in respect to the weight of the opposite force applied by the car, we equate the braking force with acceleration taken from the formula:
${{v}^{2}}={{u}^{2}}+2aS$
where u is the initial velocity, v is the final velocity, a is the acceleration of the car and S is the distance car moved after brake is applied and then equating it with the weight/force of the car.

Complete step by step solution:
To find the value of the acceleration which is taken as uniform because time is not given and so is initial speed thus making the motion of the car constant and increasing. We use the formula of the one-dimension kinematics where acceleration a, is:
$a=\dfrac{({{v}^{2}}-{{u}^{2}})}{2S}$
Placing the values of the initial velocity as 7 and final velocity as 0 and the distance travelled after the brakes are applied as $10m$. Now the acceleration of the car is:
$a=\dfrac{({{0}^{2}}-{{7}^{2}})}{2\times 10}$
$\Rightarrow a=\dfrac{({{0}^{2}}-{{7}^{2}})}{2\times 10}$
$\Rightarrow a=-2.45\text{ }m/{{s}^{2}}$
Now we know the acceleration achieved after braking while the acceleration due to the gravity is taken as $9.8\text{ }m/{{s}^{2}}$. Dividing the acceleration due to braking and the value of the acceleration due to the gravity of the car as:
$\Rightarrow \dfrac{Ac{{c}_{car}}}{Ac{{c}_{gravity}}}$
Placing the value of the acceleration ratio of both the car and the acceleration of the gravity as:
$\dfrac{Ac{{c}_{car}}}{Ac{{c}_{gravity}}}=\left| -\dfrac{2.45}{9.8} \right|$
$\Rightarrow \dfrac{Ac{{c}_{car}}}{Ac{{c}_{gravity}}}=\dfrac{1}{4}$
Placing the values of the acceleration in the ratio of the force of the car in term of acceleration due to the motion of the car and due to gravity as:
$\Rightarrow \dfrac{{{F}_{car}}}{{{F}_{gravity}}}=\dfrac{mass\times Ac{{c}_{car}}}{mass\times Ac{{c}_{gravity}}}$
$\Rightarrow \dfrac{{{F}_{car}}}{{{F}_{gravity}}} = \dfrac{mass\times 1}{mass\times 4}$
$\Rightarrow {{F}_{gravity}}=4{{F}_{car}}$

Therefore, the acceleration when brakes are on is $\dfrac{1}{4}$ of that of the weight of the car.

Note: We will use the kinematic formula ${{v}^{2}}\text{ }=\text{ }{{u}^{2}}\text{ }+\text{ }2aS$ to find the acceleration and not $S=ut+\dfrac{1}{2}a{{t}^{2}}$ as the time is not given and apart from that the car uses uniform acceleration and the acceleration ratio is negative due to the braking of the car which puts the acceleration in backward motion.