A car came to rest after 300m due to the application of brakes with re

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

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»

Here,

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

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