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A 2000 kg car traveling at 30 m/s slams on its brakes to stop. If the maximum braking force of the car is 20,000 Newtons, how far does the car travel from the time the brakes are applied until it stops?

Answer :

Final answer:
To calculate the stopping distance of the 2000kg car traveling at 30m/s with a maximum braking force of 20,000 Newtons, the work-energy principle is applied, yielding a stopping distance of 45 meters.
Explanation:
To calculate the stopping distance of the car, we use the work-energy principle. The work done by the braking force is equal to the change in kinetic energy of the car. The initial kinetic energy (KE) of the car can be determined by the formula:
KE = rac{1}{2}mv^2
where m is the mass of the car and v is the velocity. The work done by the braking force (W) is the force times the stopping distance (d), i.e. W = F imes d.
We set the work done equal to the initial kinetic energy to solve for d:
W = KE
F imes d = rac{1}{2}mv^2
Now we can rearrange the formula to solve for d:
d = rac{rac{1}{2}mv^2}{F}
By plugging in the values provided (m = 2000kg, v = 30m/s, and F = 20,000N), the calculation for stopping distance is straightforward:
d = rac{0.5 imes 2000kg imes (30m/s)^2}{20,000N}
After calculating, we find that the stopping distance is 45 meters.

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Rewritten by : Barada

Answer:

45 m

Explanation:

From the question,

The kinetic Energy of the car = Work done by the brake that is needed to stop the car.

1/2mv² = F×d......................... Equation 1

Where m = mass of the car, v = velocity of the car, F = braking force of the car, d = distance the car travels before it comes to rest.

make d the subject of the equation

d = 1/2mv²/F.......................... Equation 2

Given: m = 2000 kg, v = 30 m/s, F = 20000 N

Substitute into equation 2

d = 1/2(2000)(30²)/20000

d = 1000(30²)/20000

d = 900000/20000

d = 45 m.

Hence the car travels 45 m before it comes to rest