Question

A car moving with a speed of 40 m/s comes to rest
after travelling a distance 400 m on applying the
brakes .The acceleration of car is

Answer 1

Given that, a car moving with a speed of 40 m/s comes to rest on applying the

brakes. (Initially car is moving with a speed of 40 m/s and when brakes are applied, the car comes to rest means the final velocity of the car is 0 m/s.)

Also given that, the distance covered by the car is 400 m.

We have to find the acceleration of the car.

From the above data we have, u = 40 m/s, v = 0 m/s and s = 400 m.

Using the Third Equation Of Motion,

v² - u² = 2as

Substitute the known values in the above formula to find the value of acceleration (a).

→ (0)² - (40)² = 2(a)(400)

→ 0 - 1600 = 800a

→ -1600 = 800a

Divide by 800 on both side,

→ -1600/800 = 800a/800

→ -2 = a

(negative sign shows retardation)

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

Answer 2

In the above question , the following information is given -

A car moving with a speed of 40 m/s comes to rest after travelling a distance 400 m on applying the brakes .

Solution -

Here , the innitial Velocity of the car is 40 m / s.

So , u = 40 m / s.

The car stops finally .

Hence , the final velocity of the car is 0 m / s .

Now , according to the first equation of motion -

Acceleration = ∆ V / t

Acceleration = ( v - u ) / t

=> Acceleration = ( -40 / t )

According to the second equation of motion ,

S = ut + ( 1 / 2 ) a t ^ 2

Now ,

S refers to the displacement and is equal to 400 metres .

Substituting the given Values -

40t + ( 1 / 2 ) × ( - 40 / t ) A× t^2 = 400

=> 40t - 20t = 400

=> 20t = 400

=> t = 20 seconds. .

Substituting this value -

Acceleration = ∆ V / t

=> ( -40 / 20 ) m s^ ( -2 )

=> -2 m s^ ( -2 )

The negative sign shows retardation as the car is finally stopping ...

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