Answer:
The car will take 4 seconds.
Explanation:
First of all, we convert the power from hp to W:
[tex]40.0hp=29840W[/tex]
The work done by the car is equal to the change in its mechanical energy. This means that:
[tex]P=\frac{W}{\Delta t}=\frac{\Delta E}{\Delta t}\implies \Delta t=\frac{\Delta E}{P}[/tex]
Now, the change in mechanical energy in terms of the kinetic and potential energy is:
[tex]\Delta E= \frac{1}{2} mv_f^{2} +mgh_f-\frac{1}{2} mv_o^{2} +mgh_o[/tex]
Since the initial height and the initial velocity are zero, we have:
[tex]\Delta E= \frac{1}{2} mv_f^{2} +mgh_f[/tex]
Finally,
[tex]\Delta t=\frac{\frac{1}{2} mv_f^{2} +mgh_f}{P}[/tex]
Plugging in the given values, we obtain:
[tex]\Delta t=\frac{\frac{1}{2} (850kg)(15.0m/s)^{2} +(850kg)(9.8m/s^{2})(3.0m)}{29840W}=4.0s[/tex]
In words, the car will take 4 seconds.
