Question

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

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

________________________

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

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

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

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

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

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

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

• 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 ]

___________________________

Answer 2

Answer:

Given that :-

• Initial velocity (u) = 108 m/s

• Final velocity (v) = 0 m/s

• Time taken (t) = 4 s

• Mass (m) = 1000kg

Need to Find :-

force exerted by the brakes on the motorcar if its mass along with the passengers is 1000 kg

Solution :-

Here, at first we will find acceleration by using Newton first Equation and then we will use the formula for finding Force applied

So,

$\fbox{v = u \: + at}$

V is The Final Velocity

U is the Initial Velocity

A is the Acceleration

T is the Time

0 = 108 + a(4)

0 - 108 = a(4)

-108 = a(4)

-108 = 4a

-108/4 = a

-27 m/s = a

Hence, Retardation of car is -27 m/s²

Now,

$\fbox{f = ma}$

F is the Force

M is the Mass

A is the Acceleration

F = 1000 × -27

F = -2700 N

Hence, The force exerted by brakes is -2700 N

Know More :-

• SI unit of Acceleration is m/s²
• To change km/hr into m/s we multiply the km/hr × 5/18
• SI unit of Force is Newton
• SI unit of Mass is Kilograms
• Force is the product of mass and Acceleration
• When acceleration comes in negative it is said to be retardation