Question

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An automobile travelling with a speesd of 72km/h can be stopped within a distance of 30m,by applying brakes.What will be the stopping distance ,if the automobile speed is increased √3 times and the same braking force is applied?

Answer 1

Answer:

90 m is the new stopping distance.

Answer 2

Answer:

90m

Explanation:

Given Initial Velocity u=72kmh

= 20ms

, Distance s=30m, Final Velocity v=0,

Let acceleration is a.

Using v

2

=u

2

−2as ⇒0=20

2

+2×a×30 ⇒a=−

3

20

,

Now initial velocity is increases by

3

times and the same braking force is applied.

Let distance covered is s.

Using v

2

=u

2

−2as ⇒0=(20

3

)

2

+2×(−

3

20

)×s ⇒s=90m

Distance covered is 90 m.