## How long will it take an 850-kg car with a useful power output of 40.0 hp (1 hp equals 746 W) to reach a speed of 15.0 m/s, neglecting frict

Question

How long will it take an 850-kg car with a useful power output of 40.0 hp (1 hp equals 746 W) to reach a speed of 15.0 m/s, neglecting friction if the car also climbs a 3.00 m high hill in the pro?

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1 year 2021-09-05T08:21:07+00:00 1 Answers 69 views 0

The car will take 4 seconds.

Explanation:

First of all, we convert the power from hp to W:

$$40.0hp=29840W$$

The work done by the car is equal to the change in its mechanical energy. This means that:

$$P=\frac{W}{\Delta t}=\frac{\Delta E}{\Delta t}\implies \Delta t=\frac{\Delta E}{P}$$

Now, the change in mechanical energy in terms of the kinetic and potential energy is:

$$\Delta E= \frac{1}{2} mv_f^{2} +mgh_f-\frac{1}{2} mv_o^{2} +mgh_o$$

Since the initial height and the initial velocity are zero, we have:

$$\Delta E= \frac{1}{2} mv_f^{2} +mgh_f$$

Finally,

$$\Delta t=\frac{\frac{1}{2} mv_f^{2} +mgh_f}{P}$$

Plugging in the given values, we obtain:

$$\Delta t=\frac{\frac{1}{2} (850kg)(15.0m/s)^{2} +(850kg)(9.8m/s^{2})(3.0m)}{29840W}=4.0s$$

In words, the car will take 4 seconds.