What physics will you use to find the speed?

[deanine3]

Homework Statement

Suppose a car starts from rest and rolls down a hill that is 50 m high. You are going to find the speed of the car at the bottom of the hill. What physics will you use to find the speed? Now find the speed.

Homework Equations

KE= 1/2mv^2
PE= mgh
Speed= distance x time

The Attempt at a Solution

I am not sure where to start since there is so little information. 50m (height) x 9.8 m/s^2 (gravity)= 500 m^2/s^ but is this PE? KE? I don't have mass so I don't think its PE, and I'm sure it's not KE.
What I know: its initial velocity is 0, height is 50m, since it is "falling" it is under the influence of gravity 9.8 m/s^2.
If it is falling 50 m at 9.8 m/s that's 5.102 seconds "falling"? So the speed is 5.102s x 50m = 255.01 m/s? Is this correct? Am I along the right lines?

 

[RedBarchetta]

This is a conservation problem. Choose the bottom of the hill as h=0. This will make solving for the unknown easy, even though you can choose h=0 at the top if you wish.

So, initially the car is at rest, so it has zero kinetic energy. Since I took h=0 at the bottom, the car has gravitational potential energy at the top. At the bottom, the car will have some final kinetic energy and zero gravitational potential energy since h=0 there.

So simply set up your equations and solve for v:

<br /> \begin{gathered}<br /> K_o + U_o = K_f + U_f \hfill \\<br /> U_g = K_f \hfill \\<br /> mgh = \tfrac{1}<br /> {2}mv^2 \hfill \\<br /> gh = \tfrac{1}<br /> {2}v^2 \hfill \\<br /> v = \sqrt {2(9.8\tfrac{m}<br /> {{s^2 }})(50m)} \hfill \\<br /> v \approx 31.3\tfrac{m}<br /> {s} \hfill \\ <br /> \end{gathered} <br />

 

[deanine3]

I think I am getting however, how do you take out the 1/2 from the v^2 side of the equation? I don't see where it went.

 

[deanine3]

or are both sides multiplied by 2?

 

[RedBarchetta]

or are both sides multiplied by 2?

Yes. Then you are taking the square root of 2gh to get v.

 

[deanine3]

Ok, I really think I have it! When you divide by 2, it becomes 2gh=v^2. The v^2 becomes v=square root sign. Then, under the square root sign you plug in the 2 (from the previous division), gravity, and the height! This is slowly making sense.