Question

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Answer 1

Correct 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.

Given:-

• Mass of the motor car along with the passengers (M) = 1000 kg
• Time taken (t) = 4 s
• Initial velocity (u) = 108 km/h

To find:-

• The exerted force by the brakes on the motor car.

Solution:-

• At first , convert initial velocity (u) in m/s.
• $\implies\sf{u=108\times\dfrac{5}{18}\:m/s}$
• $\implies\sf{u=30\:m/s}$
• Final velocity (v) = 0 m/s
• Now find the exerted force (F) by the brakes on the motor car.

We know that,

${\boxed{\sf{F=\dfrac{m(v-u)}{t}}}}$

[ Put values]

$\implies\sf{F=\dfrac{1000(0-30)}{4}}$

$\implies\sf{F=\dfrac{1000\times\:-30}{4}}$

$\implies\sf{F=-7500}$

★ Opposing force is denoted by negative sign.

• Therefore, the exerted force by the brakes on the motor car is 7500 N.

Answer 2

Answer:

u = 108 km/h => 108×5/18 => 6×5 => 30 m/s

v = 0m/s

t = 4sec

therefore , a = (v-u)/t

=>a = (0-30)/4

=> a = -30/4 => -7.5 m/s²

force = mass × acceleration

force = 1000 kg × (-7.5) m/s²

force = -7500 kg m/s²

=> force = -7500 N

negative sign denotes opposing force

hope this helps you