Question

a car moving with a speed of 40 metre per second and comes at rest after traveling a distance of 400 metre on applying the breaks . what is its acceleration​

Answer 1

Explanation:

u= 40m/s

v=0m/s

s=400m

v^2-u^2=2as

02-(40)2=2×a×400

-1600=800a

a=-1600/800

=-2m/s2

Answer 2

Given that, a car is moving with a speed of 40 m/s (initial speed i.e. u of the car is 40 m/s). The car comes at rest after traveling a distance of 400 metre on applying the breaks.

We have to find the acceleration of the car.

From above data we have, initial velocity/speed of the car = 40 m/s, final velocity of the car = 0 m/s (as brakes are applied) and distance i.e. s covered by the car is 400 m.

First equation of motion:

v = u + at

Second equation of motion:

s = ut + 1/2 at²

Third equation of motion:

v² - u² = 2as

We have value of v, u, and s. So,

Using the Third Equation Of Motion,

v² - u² = 2as

Substitute the known in the above formula,

→ (0)² - (40)² = 2(a)(400)

→ 0 - 1600 = 800a

→ -16 = 8a

Divide by 8 on both sides

→ -16/8 = 8a/8

→ -2 = a

Therefore, the acceleration of the car is -2 m/s. (Negative sign shows retardation).