Question

QUESTION:-
A car moving with a speed of 40 m/s is brought to rest be application of brakes in 400 m.Calculated the acc. and time taken

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Answer 1

Answer:

-2 m/s^2 acc.

tmie taken 20 seconds

Explanation:

for acceleration,

we know that,

Vf^2 = Vi^2 + 2ad

a= Vf^2 - Vi^2 / 2d

= -1600/800

= - 2 m/s^2

for calculating time taken ,

we know that,

Vf = Vi + a*t

t = Vf - Vi / a

= 0 - 40 / -2

= 20 sec.

Answer 2

Answer:

A car is moving with a speed of 40 m/s. On applying brake, it comes to rest at 20 s. What is the distance between the distance covered by the car before it stops?

Given

initial velocity (u) = 40 m/s

Final velocity ( V) = 0 m/s

Time taken ( T ) = 20 second

now first we find the retardation after then we find distance

V= u + at

or, a =( V- u)/T

or a = -40/20

or a = - 2 m/s²

now V² = u² + 2as

or S = - (40)²/ 2 (-2)

or S = 400 m

hence distance coveres before come to rest is 400 m