How Long Does It Take a Car to Climb a Hill with Limited Power?

[Sunbodi]

Homework Statement

Your 1500-kg car, moving at 7.0 m/s, approaches the bottom of a hill that is 20 m high (Figure 1) . To save gas, you use on average only 3.2kW of engine power, realizing that half of the energy delivered by the engine and half of the initial kinetic energy will be dissipated. If you calculated correctly, your car will just barely make it over the hill.

What time interval is required for your car to travel from the bottom to the top of the hill?

Homework Equations

KE=1/2mv^2
PE=mgh
P=W/T
y=v0t+.5at^2

The Attempt at a Solution

KE=.5(1500)(7)^2
=36700 Joules or 36.75 kJ
P=W/T
3.2 KW=36.75 KJ/T
T=11.48 SecondsIT SAYS I'M WRONG SO I DID THIS

"half of the initial kinetic energy will be dissipated"
3.2=18.38/T
T=5.74

IT STILL SAY I WRONG SO I DO THIS
20=7t+.5(9.8)t^s

t=1.42 seconds (Used quadratic formula to solve for T).

But then I think perhaps the car is accelerating I mean that's the only way it has power right? So the acceleration is not 9.8 but 9.8 plus something because the acceleration of the car has to overcome the acceleration due to gravity in order to make it over the hill. I tried substitution of the time in P=W/T into the y=v0t+.5at^2 equation in order to find the acceleration of the car but that's just a pain in the ass and you know what my graphing calculator told me? It told me the acceleration is 9.8 I kid you not. I can't get the right answer and I feel so dumb and I know I'm so dumb please help.

 

[TomHart]

The problem states that the car just barely makes it over the hill. That means that the car has slowed down from the time it started until it reached the top. And since it "just barely makes it", you can interpret that to mean that the speed is effectively 0.00 m/s when it reaches the top. So the car is decelerating.

So one thing I would want to know is how much energy is required for the car to get to the top of the hill. In other words, how much does its potential energy have to increase from the bottom of the hill to the top of the hill. Then, how does that compare to the initial kinetic energy of the car. But there's a hitch. See next paragraph. By the way, your initial kinetic energy calculation looks correct.

One thing that adds difficulty to the problem is that it states that half of the engine energy and half of the kinetic energy is dissipated. Where does it go? I don't know. Maybe it is a very rough, sticky road surface. But you have to take into account that only half of those sources of energy gets converted to potential energy. The other half is lost - converted to heat, basically.

So since half of the kinetic energy is lost, you only have half remaining. Is that enough to get the car to the top of the hill? If not, then where does the car get the rest of the energy? Well, it has to come from the engine. But remember, half of it is lost too.

Hang in there. If you work hard and don't give up, you can be successful at this.

 

[haruspex]

y=v0t+.5at^2

That equation is only valid for constant acceleration. The shape of the hill is uncertain, the engine power is variable - only the average is known - so you cannot assume any particular acceleration profile.
Suppose it takes time t. As TomHart suggests, concentrate on the work done.