A 1,500 kg car is traveling at 25 m/s. The driver suddenly applies the brakes causing the car to skids to a stop. If the average braking force between the tires and the road is 7,100 N, how far does the car slide before it comes to rest?​

Respuesta :

By Newton's second law, the car slows down with an average acceleration a such that

-7100 N = (1500 kg) a   ⇒   a ≈ -4.7 m/s²

If we treat the car as accelerating uniformly with this magnitude, then the car slides a distance ∆x such that

0² - (25 m/s)² = 2a ∆x   ⇒   ∆x ≈ 66 m

The car slide before it comes to rest will be 66 m. Newton's second equation of motion is applied in this problem.

What is force?

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Force is defined as the product of mass and acceleration. Its unit is Newton.

The acceleration is found as;

F=ma

-7100 N = (1500 kg) a

a= -4.7 m/s²

If we assume that the automobile accelerates uniformly with this magnitude, the car slides x distance. From the Newton's second equation of motion;

v²=u²+2as

v²-u²=2as

0² - (25 m/s)² =2×(-4.7) s  

s=66 m

Hence,the car slide before it comes to rest will be 66 m.

To learn more about the force, refer to the link;

https://brainly.com/question/26115859

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