3) a driver in a 1000-kg car traveling at 24 m/s slams on the brakes and skids to a stop. if the coefticient of friction between the tires and the level road is 0 80 how long will the skid marks be? a) 37 m b) 30 m c) 34 m d 46 m

The length of skidding marks will be of 37 m. Hence, option (A) is correct.

Given data:

The mass of car is, .

The initial speed of car is, u = 24 m/s.

The final speed of car is, v=0.  (Car stops finally due to friction)

The coefficient of friction between the tires road is,  .

The linear force acting on the car is opposed by the frictional force. Therefore,

Linear force = frictional force

Now, apply the third kinematic equation of motion as,

Thus, the length of skidding marks will be of 37 m. Hence, option (A) is correct.

Learn more about the frictional force here:

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A) 37 m

Explanation:

The car is moving of uniformly accelerated motion, so the distance it covers can be calculated by using the following SUVAT equation:

(1)

where

v = 0 m/s is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d is the length of the skid

We need to find the acceleration first. We know that the force responsible for the (de)celeration is the force of friction, so:

where

m = 1000 kg is the mass of the car

is the coefficient of friction

a is the deceleration of the car

g = 9.8 m/s^2 is the acceleration due to gravity

The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:

And we can substitute it into eq.(1) to find d:

the answer would be 32*f