Question

Q. A motorcar is moving with a velocity of 108 m/s and it takes 4 s to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is 1000 kg
Plz ans this ...​

Answer 1

Given

Initial velocity (u) = 108 m/s

Final velocity (v) = 0 [ Brakes applied ]

Time taken (t) = 4 s

Mass (m) = 1000kg

To find

The force exerted by the brakes on the motorcar

Solution

According to the first equation of motion

$\\ \: \: \: \: \bigstar\large{\purple{\underline{\boxed{\bf{\purple{v= u + at}}}}}}$

Substitute all the values

$\\\implies\tt 0=108+a\times{4}$

$\implies\tt 0=108+4a$

$\implies\tt -108 = 4a$

$\implies\tt a=\cancel\dfrac{-108}{4}$

$\implies\tt a= -27m/s^2$

$\therefore{\tt{\purple{Acceleration\:of\:motorcar=-27m/s^2}}}$

$\large{\underline{\underline{\bf{\red{Note:}}}}}$

Minus shows retardation of motor car

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$\\{\underline{\tt{\purple{Let's\:find\:out\:the\:force\: exerted\:by\:brakes}}}}$

$\bigstar{\boxed{\bf{Force=mass\times acceleration}}}$

$\implies\tt F=m\times{a}$

$\implies\tt F=1000\times{(-27)}$

$\implies\tt F=-2700\:N$

$\therefore{\tt{\red{Force\:exerted\:by\:the\:brakes\:on\:motorcar=-2700N}}}$

$\large{\underline{\underline{\bf{\purple{Note:}}}}}$

Minus shows force exerted in opposite direction

$\\\large{\underline{\underline{\sf{\red{Basic\:Information}}}}}$

The rate of change in velocity is known as Acceleration

s = ut + ½ at² [ second equation of motion ]

v² = u² + 2as [ Third equation of motion ]

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Answer 2

Step-by-step explanation:

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