Question

$\huge \bold{quєѕtíσn}$
A motorcar is moving with a velocity of 108 km/h 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.


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Answer 1

Step-by-step explanation:

Given that

$Initial \: velocity \: of \: motor \: car = 108 km/h$

$⇒ (u) = 108×100060×60108×100060×60 = 30 m/s$

$Final velocity (v) = 0 m/s$

$Total mass of the motor car along with its passengers (m) = 1000 kg.$

$Time \: is \: taken \: to \: stop \: the \: motor \: car \: (t) \: = \: 4 \: sec.$

$using \: f = m (v - u)/t$

$= 1000(0 - 30)/4$

$= - 7500 N$

The negative sign shows that the force exerted by the brakes is opposite to the direction of motion of the motor car.

Answer 2

Answer:

$\tt \underline \red{•QuEstIOn:-}$

A motorcar is moving with a velocity of 108 km/h 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.

$\tt \underline \pink{•AnSwER:-}$

→We are given that Initial velocity of motor car = 108 km/h

$\tt \: \rightarrow(u) = \frac{108 \times 1000}{60 \times 60} = 30 \: m/s$

→Final velocity (v) =0 m/s

→total mass of motor car along with its passengers (m)= 1000 kg.

→time is taken to stop the motor car (t) = 4sec.

$\tt \: =using \: F = m( \frac{v - u}{t} )$

$\tt = \frac{1000(0 - 30)}{4} = - 7500kg.m/ {s}^{2}$

$\tt = - 7500 \: N$

The negative sign shows that the force exerted by the brakes is opposite to the direction of motion of the motor car.