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How long (in s) will it take an 875 kg car with a useful power output of 42.0 hp (1 hp = 746 W) to reach a speed of 17.0 m/s, neglecting friction? (Assume the car starts from rest.)

Answer :

Answer:

The time taken by the car with a useful power output is 4.03 seconds.

Explanation:

Given that,

Mass of the car, m = 875 kg

Initial speed of the car, u = 0

Final speed of the car, v = 17 m/s  

Power of the car, P = 42 hp = 31332 W

The power output of the car is given by the work done per unit time. It is given by :

[tex]P=\dfrac{W}{t}[/tex]

Initial kinetic energy of the car will be 0 as it is at rest.

Final kinetic energy of the car is :

[tex]K=\dfrac{1}{2}mv^2[/tex]

[tex]K=\dfrac{1}{2}\times 875\times (17)^2[/tex]

K = 126437.5

As per the work energy theorem, the change in kinetic energy of the object is equal to the work done.

So,

[tex]P=\dfrac{W}{t}[/tex]

[tex]t=\dfrac{W}{P}[/tex]

[tex]t=\dfrac{126437.5}{31332}[/tex]

t = 4.03 seconds

So, the time taken by the car with a useful power output is 4.03 seconds. Hence, this is the required solution.

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