Question

6. A motor car of mass 2000 kg is moving with
a certain velocity. It is brought to rest by the
application of brakes, within a distance of
20 m when the average resistance being
offered to it is 5000 N. What was the velocity
of the motor car?
​

Answer 1

Answer :-

Velocity of the motor car was 10 m/s .

Explanation :-

We have :-

→ Mass of the motor car = 2000 kg

→ Final velocity = 0 m/s

→ Distance = 20 m

→ Average resisting force = 5000 N

Since, the force acting here is resisting force which finally brings the car to rest. So, it's magnitude will be -ve i.e -5000 N.

________________________________

Firstly, we will calculate the acceleration of the car by using Newton's 2nd law of motion.

F = ma

⇒ -5000 = 2000(a)

⇒ a = -5000/2000

⇒ a = -2.5 m/s²

Now, we will calculate initial velocity of the car by using the 3rd equation of motion.

v² - u² = 2as

⇒ 0 - u² = 2(-2.5)(20)

⇒ -u² = -100

⇒ u² = 100

⇒ u = √100

⇒ u = 10 m/s

Answer 2

Answer:

Given :-

• A motor car of mass 2000 kg is moving with a certain velocity. It is brought to rest by the application of brakes, within a distance of 20 m when the average resistance being offered to it is 5000 N.

To Find :-

• What is the velocity of the motor car.

Formula Used :-

$\clubsuit$ Force Formula :

$\longmapsto \sf\boxed{\bold{\pink{F = ma}}}$

where,

• F = Force
• m = Mass
• a = Acceleration

$\clubsuit$ 3rd Equation of Motion :

$\longmapsto \sf\boxed{\bold{\pink{v^2 - u^2 =\: 2as}}}$

where,

• v = Velocity
• u = Initial Velocity
• a = Acceleration
• s = Distance Travelled

Solution :-

First, we have to find the acceleration :

Given :

$\bigstar$ Force (F) = - 5000 N

$\bigstar$ Mass (m) = 2000 kg

According to the question by using the formula we get,

$\dashrightarrow \sf\bold{\purple{F =\: ma}}$

$\mapsto \sf\bold{a =\: \dfrac{F}{m}}$

$\implies \sf a =\: \dfrac{- 5\cancel{000}}{2\cancel{000}}$

$\implies \sf a =\: \dfrac{- 5}{2}$

$\implies \sf\bold{\green{a =\: - 2.5\: m/s^2}}$

Hence, the acceleration is - 2.5 m/s².

Now, we have to find the velocity :

Given :

$\bigstar$ Final Velocity (v) = 0 m/s

$\bigstar$ Acceleration (a) = - 2.5 m/s²

$\bigstar$ Distance travelled = 20 m

According to the question by using the formula we get,

$\longrightarrow \sf (0)^2 - u^2 =\: 2 \times (- 2.5) \times 20$

$\longrightarrow \sf 0 - u^2 =\: 2 \times \bigg(- \dfrac{25}{10}\bigg) \times 20$

$\longrightarrow \sf 0 - u^2 =\: 2 \times \bigg(- \dfrac{50\cancel{0}}{1\cancel{0}}\bigg)$

$\longrightarrow \sf 0 - u^2 =\: 2 \times \bigg(- \dfrac{50}{1}\bigg)$

$\longrightarrow \sf 0 - u^2 =\: 2 \times (- 50)$

$\longrightarrow \sf 0 - u^2 =\: - 100$

$\longrightarrow \sf {\cancel{-}} u^2 =\: {\cancel{-}} 100$

$\longrightarrow \sf u^2 =\: 100$

$\longrightarrow \sf u =\: \sqrt{100}$

$\longrightarrow \sf\bold{\red{u =\: 10\: m/s}}$

$\therefore$ The velocity of the motor car is 10 m/s .