Question

A car came to rest after 300m due to the application of brakes with retardation of 4 m/s2 . The velocity of the car before applying the brakes is ..........

Answer 1

Given:-

• A car came to rest.[ v = 0 ]
• And it travelled 300m and stopped and it has an retardation of 4m/s²

To Find:-

• Find the initial velocity of the car.

Solution:-

Here,

• u = initial velocity
• v = final velocity
• a = acceleration
• s = distance
• t = time

We can find the initial velocity by using second equation of motion:-

u = ? ; v = 0 ; s = 300m ; a = -4m/s²

➣ v² - u² = 2as

➣ 0² - u² = 2(-4)(300)

➣ -u² = -2400

➣ u = √2400

➣ $\boxed{\bf{\color{yellw}{ \: u \: = \: 49m/s \: }}}$

Answer 2

Answer :-

Given :-

• Distance travelled, s = 300 m
• Acceleration, a = - 4 m/s² ( negative sign due to retardation )
• Final velocity, v = 0 m/s ( comes to rest )

To Find :-

• Initial velocity, u

Solution :-

Substituting the values in 3rd equation of motion -

→ v² = u² + 2as

→ 0 = u² + 2 × (-4) × 300

→ u² = 8 × 300

→ u² = 2400

→ u² ≈ 49

Initial velocity = 49 m/s