Answer
77.7k+ views
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 ${\text{ = }}$ ${{Mass \times acceleration}}$
${\text{F = }}\;{{m \times a}}$.
Complete step by step answer:
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.
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 ${\text{ = }}$ ${{Mass \times acceleration}}$
${\text{F = }}\;{{m \times a}}$.
Complete step by step answer:
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.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write an article on the need and importance of sports class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Choose the one which best expresses the meaning of class 9 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)