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A 1200-kg car, initially moving at 20 m/s, comes to a stop at a red light over a time of 3 s from the moment the driver hit the brakes. What is the impulse delivered to the car by the static friction force (assumed constant) between the road and the tires

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Answer:

Explanation:

Mass of car (M)=1200kg

Initial velocity (u)=20m/s

Stop after time (t)=3sec.

Come to stop implies that the final velocity is zero, v=0m/s

Using newton second law of motion

F=m(v-u)/t

Ft=m(v-u)

Since impulse is Ft

I=Ft

Then, I=Ft=m(v-u)

I=m(v-u)

I=1200(0-20)

I=1200×-20

I=-24,000Ns

The impulse delivered to the car by static friction is -24,000Ns

The  impulse delivered to the car by static friction is -24,000Ns

Calculation of the impulse:

Since A 1200-kg car, initially moving at 20 m/s, comes to a stop at a red light over a time of 3 s from the moment the driver hit the brakes.

Now here we used second law of motion of newton.

F=m(v-u)/t

Ft=m(v-u)

Since impulse is Ft

So,

I=Ft

Now

, I=Ft=m(v-u)

So,

I=m(v-u)

I=1200(0-20)

I=1200×-20

I=-24,000Ns

Learn more about force here: https://brainly.com/question/3398162

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