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# A car going at a speed of ${\text{7m}}{{\text{s}}^{{\text{ - 1}}}}$. Can be stopped by applying brakes in $\alpha$ shortest distance of $10$m. Show that the total friction force opposing the motion, when brakes are applied. Is $1/{4^{th}}$ of the weight of the car. $\left( {{\text{g = }}\;{\text{9}}{\text{.8m}}{{\text{s}}^{{\text{ - 1}}}}} \right)$

Last updated date: 21st Apr 2024
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Hint: By seeing the question we can say that if the car stopped its final velocity will become $0$, Hence by given data we can calculate acceleration. Also the mass of the car is not given in the question but we know that force is mass multiplied by product. From force we can find mass.

Formula used:
$\left( {\text{i}} \right)\;{{\text{V}}^{\text{2}}}\;{\text{ = }}{{\text{u}}^{\text{2}}}{\text{ + 2as}}$
$V$ is final velocity
$u$ is initial velocity
$a$ is acceleration
$\left( {{\text{ii}}} \right)$ Force ${\text{ = }}$ ${{Mass \times acceleration}}$
${\text{F = }}\;{{m \times a}}$.

We have given, a car is at speed ${\text{7m}}{{\text{s}}^{{\text{ - 1}}}}$ moving and suddenly it stopped by applying brakes so its final velocity become $0.$ and distance travelled by it is ${\text{10m}}$. We have to show the total resistance force (functional force) is $\dfrac{1}{4}$ times the weight of the car.
Data given,
${\text{u = }}\;{\text{7m/s}}\;{\text{,V = 0,}}\;{\text{s = 10m,}}\;{\text{& }}\;{\text{acceleration = }}\;{\text{?}}$
So,
By using ${{\text{3}}^{{\text{rd}}}}$ equation of Motion
${\text{V = }}\;{{\text{u}}^{\text{2}}}{\text{ + 2as}}$
So, ${\text{a = }}\;\dfrac{{{{\text{V}}^{{2}}}{\text{ - }}{{\text{u}}^{\text{2}}}}}{{{\text{2s}}}}$
${\text{a}}\;{\text{ = }}\;\dfrac{{{{\left( {\text{0}} \right)}^{\text{2}}}{\text{ - }}{{\left( {\text{7}} \right)}^{\text{2}}}}}{{{{2 \times 10}}}}$
$\Rightarrow {\text{a}}\;{\text{ = }}\;\dfrac{{{\text{ - 49}}}}{{{\text{20}}\;}}\;\; \Rightarrow {\text{ - 2}}{\text{.45m/}}{{\text{s}}^{\text{2}}}$
Now,
$2.45$ can also be written as $a = \;\dfrac{{49}}{{20}}$
$\Rightarrow \; a = \dfrac{{49 \times 2}}{{20 \times 2}}\; = \;\dfrac{{98}}{{40}}\; = \;\dfrac{{9.8}}{4}\; = \dfrac{g}{4}\;\left( {\because g = \,9.8} \right)$
Now, the resistance or frictional force ${\text{ = }}\;{\text{ma}}$

So, the total frictional force opposing the motion in $\dfrac{1}{4}$ times the weight of car R(functional force) $= \;\dfrac{1}{4}$ weight of car.

Note: In order to solve this question, we need to count the acceleration in terms of ${\text{g}}$ as you have seen in the solution. Negative sign shows that the motion is in the opposite direction with the frictional force or resistance force. It must be noted that frictional force is always in the opposite direction of the motion.