Question

A mestos car is moving with a
mawing with a velocity of 108 km/h .
and it deckes 4
second to stop after the brakes are
appuid. calculates the force exerted by the brakes on they
motor car if its mass along also the parengers to 100kg.
Teacher's Signature
REMARK
VARDHMAN
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Answer 1

Given:-

• Initial velocity ,u = 108km/h = 108×5/18 =30m/s
• Final velocity ,v = 0m/s
• Time taken ,t = 4
• Mass ,m = 100kg

To Find:-

• Force exerted by Brake ,F

Solution:-

Firstly we calculate the acceleration of the car

Acceleration is defined as the rate of change of velocity at per unit time.

• a = v-u/t

where,

v is the final velocity

a is the acceleration

u is the initial velocity

t is the time taken

Substitute the value we get

→ a = 0-30/4

→ a = -30/4

→ a = -7.5 m/s²

Here negative sign show retardation .

Now, Calculating the Force exerted by Brake on the car.

Force is defined as the product of mass and acceleration.

• F = ma

Where,

F is the Force

m is the mass

a is the acceleration

Substitute the value we get

→ F = 100 × (-7.5)

→ F = -750 N

Here negative sign show that the force is acting in opposite direction of motion.

Therefore,the force applied by the brake is 750 Newton.

Answer 2

Answer:

Given :-

• Initial velocity = 108 km/h = 30 m/s
• Final velocity = 0 m/s
• Time taken = 4 sec
• Mass = 100 kg

To Find :-

Force exerted

SoluTion :-

Firstly we will find its acceleration

$\huge \bf \: a = \frac{v - u}{t}$

Here,

A = Acceleration

V = Final Velocity

U = Initial velocity

T = Time taken.

$\sf \mapsto \: a = \dfrac{0 - 30}{4}$

$\sf \mapsto \: a = \dfrac{ - 30}{4}$

$\sf \mapsto \: a = - 7.5 \: mps {}^{2}$

Note :-

Negative sign shows Retardation

Let's find force exerted

$\huge \tt \: F \: = ma$

Here,

F = Force applied

M = Mass

A = Acceleration

$\sf \mapsto \: f = 100 \times - 7.5$

$\sf \mapsto \: F = - 750 \: n$