Question

a car is moving with a velocity of 108km/h. If the driver of the car wants to change its velocity from 108km/h to 90km/h in 2 minutes. The mass of the car is 300kg. what force applied by the brakes on the tires?

Answer 1

Given :

• Initial velocity of a car, u = 108 km/h
• Final velocity of a car, v = 90 km/h
• Time taken, t = 2 minutes
• Mass of a car, m = 300 kg

To find :

• Force applied by the brakes on the tires, F

According to the question,

After changing Initial and final velocity in m/s and time in seconds. The values are,

• Initial velocity, u = 30 m/s
• Final velocity, v = 25 m/s
• Time, t = 120 seconds

➞ v = u + at [ First equation of motion]

Where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• t = Time taken

➞ Substituting the values,

➞ 25 = 30 + a × 120

➞ 25 - 30 = 120a

➞ - 5 = 120a

➞ - 5 ÷ 120 = a

➞ - 1/24 = a

• So, the acceleration is - 1/24 m/s².

Note : Negative signs means retardation.

Now,

➞ Force = Mass × Acceleration

Or,

➞ F = ma

➞ F = 300 × ( - 1/24)

➞ F = - 12.5 N

• So, the force applied by the brakes on the tires is - 12.5 Newton.

Note : Negative signs means that the force is acting on opposite direction.

_____________________

Answer 2

Answer:

Given :-

• Initial velocity of car = 108 km/h
• Final velocity of car = 90 km/h
• Time taken = 2 minutes
• Mass of car = 300 kg

To Find :-

Force applied

Formula :-

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

Here,

V = Final Velocity

U = Initial velocity

A = Acceleration

T = Time

$\huge \fbox{F = ma}$

Here,

F = Force applied

M = Mass

A = Acceleration

Solution :-

$\tt \pink {Initial\: Velocity = 108 = 108 \times \dfrac{5}{18} = 30 \: mps}$

$\tt \pink{Final \: velocity = 90 = 90 \times \frac{5}{18} = 25 \: mps}$

$\tt \pink{ 2 \: minutes \: = 2 \times 60 = 120 \: sec}$

Now,

Let's find Acceleration

$\sf \red { \implies \: 25 = 30 + a \times 120}$

$\sf \red { \implies \: 25 = 30 + 120a}$

$\sf \red{ \implies \: 25 - 30 = 120a}$

$\sf \red{ \implies - 5 = 120a}$

$\sf \red { \implies \: a = \dfrac{ - 5}{120}}$

$\sf \red{\implies \: a = \frac{ - 1}{24} }$

Hence :-

Acceleration of car is -1/24 m/s².

$\sf \blue{F = 300 \times \dfrac{ - 1}{24} }$

$\sf \blue {F = - 12.5 \: N}$

Hence :-

Force applied is -12.5 Newton.