Question

A car travels a distance of 100m with a constant acceleration and average velocity of 20m/s .The final velocity acquired by the car is 25m/s. Find initial velocity and acceleration of the car
​

Answer 1

Required Answer :

As per the provided information in the given question, we have :

• Distance covered, s = 100 m
• Average velocity, $\overline{\nu}$ = 20 m/s
• Final velocity, v = 25 m/s

We've been asked to calculate the initial velocity and acceleration of the car.

As we know that, average velocity is given by :

⇒ $\boxed{\sf{ \overline{\nu} = \dfrac{v + u}{2} }}$

• v denotes final velocity
• u denotes initial velocity
• $\overline{\nu}$ denotes average velocity

⇒ $\overline{\nu}$ = (u + 25)/2

⇒ 20 = (u + 25)/2

⇒ 20 × 2 = u + 25

⇒ 40 = u + 25

⇒ 40 ― 25 = u

⇒ 15 m/s = u

Therefore, initial velocity of the car is 15 m/s.

Now, we need to calculate the acceleration of the car. By using the third equation of motion,

⇒ v² — u² = 2as

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance

⇒ (25)² — (15)² = 2 × a × 100

⇒ 625 ― 225 = 200a

⇒ 400 = 200a

⇒ 400 ÷ 200 = a

⇒ 2 m/s² = a

Therefore, acceleration of the car is 2 m/s².

Answer 2

QUESTION :-

A car travels a distance of 100m with a constant acceleration and average velocity of 20m/s .The final velocity acquired by the car is 25m/s. Find initial velocity and acceleration of the car

GIVEN :-

car travels a distance of 100m with a constant acceleration and average velocity of 20m/s

final velocity acquired by the car is 25m/s

TO FIND :-

Find initial velocity and acceleration of the car = ?

SOLUTION :-

Average velocity V = ( v + u ) /2

20 = (25+u)/2

u = 40 - 25

u = 15 (ms^2)

V² - u² = 2s

a = (252 - 152)/ (2× 100)

a = (625 -225)/200

a = 2 (ms^2)

initial velocity and acceleration

of the car = 2 (ms^2)