Finding velocity/time graph for a car?

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In summary, The speaker has been trying to find a velocity versus time graph for a car they are fixing, but they have run into some problems. They approximated a power versus velocity graph and looked at the formula for air resistance and rolling resistance. However, the result they got doesn't make sense and they are seeking help to verify their work. They also discuss the relationship between power, force, and velocity.
  • #1
andrewr0x
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I have been recently playing around with some figures in an attempt to find a velocity versus time graph under ideal conditions including air resistance and rolling resistance for a car that I am in the process of fixing. What better incentive to get it running than knowing (at least a rough estimate) of how fast it will be? However, I have run into some problems.

I have Approximated a Power versus velocity graph based on a graph that I got from finding a cubic regression from a dynamometer chart of the engine in the car. I then altered tha graph so that instead of having power versus rpm of the flywheel to power versus velocity given the gear ratios and radius of the tire.
v=(2pi*x/60)/(Ratio of gear * Ratio of Differential)*Radius of Tire
where v is velocity and x is the RPM of the engine.

Once this has been found, I determined that the derivative of P(v) would equal F(v) due to the engine. I also looked up the formula for F(v) of air resistance to be (1/2)*Coefficient of drag*Frontal area*Air density*velocity^2. Rolling resistance's would be Coefficient of rolling friction*Normal.

From this I get a differential equation that m(dv/dt)=Fengine(v)-Fair(v)-Ffriction(v).

However, upon solving this, I get a result that does not make sense at all. Could someone verify that I am doing this correctly, and if not, give me some instructions on how to do this the correct way? Any and all help is appreciated.
 
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  • #2
The force given by engine is not the derivative dP/dv. It is P/v.
 
  • #3
krab said:
The force given by engine is not the derivative dP/dv. It is P/v.
But I thought that the integral of Force versus Velocity gives you Power?

Can you explain why the force of an engine would be power/velocity?
 
  • #4
[tex] P = \frac{dW}{dt} [/tex]

[tex]F = \frac{dW}{dx} [/tex] so [tex] dW = F dx [/tex]

then

[tex] P = \frac{F dx}{dt} [/tex]

If F is assumed to be constant it simplifies to the very nice

[tex] P = F \frac{dx}{dt} = Fv [/tex]
 
  • #5
The thing about this is that the force of my engine is not constant. So I need a way to allow for this in my equations
 
  • #6
If the force is varying with time, just differentiate the second last expression, if it is varying with some other variable, then use the chain rule to find the net rate of change WRT to time.
 
  • #7
whozum said:
[tex] P = \frac{dW}{dt} [/tex]

[tex]F = \frac{dW}{dx} [/tex] so [tex] dW = F dx [/tex]

then

[tex] P = \frac{F dx}{dt} [/tex]

If F is assumed to be constant it simplifies to the very nice

[tex] P = F \frac{dx}{dt} = Fv [/tex]
That's all true except the last sentence. F needn't be constant.
[tex] P = \frac{F dx}{dt} = F \frac{dx}{dt} = Fv [/tex]
 
  • #8
Thank you guys, I will try and play around with that formula. I was playing around in Derive (very nice once you get the hang of it) and I got a velocity curve that made no sense at all. After thinking about that I remember reviewing that Power = Force * Average Velocity in my Physics C class.
 
  • #10
I am getting a force of 19k Newtons. This seems unreasonably high to be pulling about 1.4 G's in a 3000 pound car with only 160 hp to the wheels. What would cause this?
 
  • #11
160hp can get you this force, but at only 14 mph.
 

Related to Finding velocity/time graph for a car?

1. What is a velocity/time graph for a car?

A velocity/time graph for a car is a graphical representation of how the velocity of a car changes over time. It shows how fast the car is moving at different points in time.

2. How is a velocity/time graph for a car created?

A velocity/time graph for a car is created by plotting the velocity of the car on the y-axis and time on the x-axis. The data points are then connected with a line to show the changes in velocity over time.

3. What does the slope of a velocity/time graph for a car represent?

The slope of a velocity/time graph for a car represents the acceleration of the car. A steeper slope indicates a higher acceleration, while a flatter slope indicates a lower acceleration.

4. How can you determine the distance traveled from a velocity/time graph for a car?

The distance traveled can be determined by finding the area under the velocity/time graph. This can be done by dividing the graph into smaller shapes (such as rectangles or triangles) and calculating the area of each shape, then adding them together.

5. What can a velocity/time graph for a car tell us about the motion of the car?

A velocity/time graph for a car can tell us about the speed, direction, and acceleration of the car at different points in time. It can also show if the car is moving at a constant speed or if it is accelerating or decelerating.

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