Car Coasting Up a Hill: Equations for Velocity and Height

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SUMMARY

The discussion focuses on deriving equations for velocity and height of a car coasting up a hill with a constant slope. Key equations include kinetic energy (Ke = (1/2)MV^2) and the force acting on the car due to gravity (d(Ke)dt = -m*g*sin(slope)v). The primary challenge is to express velocity as a function of time while considering the conversion of kinetic energy to potential energy as the car ascends. The solution involves recognizing that the deceleration of the car can be described using Newton's second law, where the acceleration (a) is constant at -g*sin(∅).

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Homework Statement



This is not a homework question per se but rather something I was thinking about while driving the other day.

I am in a car on a hill with constant slope traveling at an initial speed. How can I come up with a set of equations describing both velocity as a function of time and and height as a function of time.


Homework Equations


Ke = (1/2)MV^2
d(Ke)dt = -m*g*sin(slope)v


The Attempt at a Solution



All attempts to find solutions to this question have been fruitless. Most physics / calc books seem to offer this question in a simplified version: how far up the hill does the car make it. I am far more interested in plotting the curve of speed as a function of time.

Based on my understanding:

As the car travels up the hill, kinetic energy (Ke) is converted to potential energy: Ke = Pe. Potential energy = mgh. The kink seems to be that the change in potential energy is not the same. At the start, speed is the greatest, thus the change in potential energy is greatest ( = mg(sin(slope)*v). However as the car climbs the hill, its speed changes, thus the rate at which its speed changes, changes.

so far:
Initial conditions:
Keo = (.5)*mvo2.
dKe/dt = -mgsin(∅)vo

how can I relate these to t, take a derivative of the first equation and set it equal to the 2nd?
 
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Your dKe/dt notion is not needed. -mgsin(∅)=F is, in fact, a force. It's the force slowing your car down. You can convert that into the rate at which you car is slowing down by using Newton's law, F=ma. So a=(-gsin(∅)) is the deceleration rate of your car. And it's a constant since the slope is constant. You really don't need energy at all to solve the problem you've got in mind.
 
Last edited:

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