Force due to wind and rolling friction on van

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SUMMARY

The discussion centers on calculating the force due to wind and rolling friction on a minivan coasting at a speed of 44.17 mph, with a total mass of 2500 kg. The correct approach involves determining the instantaneous acceleration by finding the derivative of the speed-time graph at the specified time of 19 seconds. The final calculated force is 764 N, which is derived from applying Newton's second law (F=ma) after obtaining the instantaneous acceleration from the graph's tangent line.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Ability to interpret speed-time graphs
  • Knowledge of calculus, specifically derivatives
  • Familiarity with units of force (Newtons)
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  • Learn how to calculate derivatives from graphs
  • Study the concept of instantaneous acceleration
  • Explore the relationship between force, mass, and acceleration in physics
  • Review examples of real-world applications of Newton's laws
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of force calculations in real-world scenarios.

fruitl00p
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Homework Statement


The graph shows the speed as a function of time for a minivan, coasting on neutral along a straight, level road, loaded with windsurfing equipment and towing a boat. The total mass is 2500 kg, weight= 5512lbs. Find the size of the force due to wind and rolling friction when the van speed is 44.17 mph. (1 mph=0.447m/s)


Homework Equations


F=ma


The Attempt at a Solution


I don't think anyone needs to see the graph to understand my problem. Well I hope.

I understand that a= delta v/delta t. But whenever I use the equation I get an acceleration that does not work. Then I thought that since the equation wants the force at 44.17 mph, that I need to find the instantaneous acceleration.

That's my problem: I don't know how to get the instantaneous acceleration. At 44.17 mph the time is at 19 s. Should I create a derivative. If so, how do I go about it?

The answer for this problem is 764 N

Please tell me why my approach is incorrect.
 
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fruitl00p said:
That's my problem: I don't know how to get the instantaneous acceleration. At 44.17 mph the time is at 19 s. Should I create a derivative. If so, how do I go about it?


That's right, calculate the instantaneous acceleration by finding the derivative of the graph at the given time. You must be able to find a way to do that (draw a tangtent).

Good luck!
 
da_willem, that is my problem. I do not know how to find the derivative of the graph. If I am given a derivative, I can do it; but to look at a graph and create one I just don't know.
 
In the graph you draw a line tangential to a point where you want to know the derivative (a 'tangent line'). The derivative in that point is now the 'slope' of this line, i.e. a vertical interval (\Delta v[/tex]) divided by a corresponding horizontal interval (\Delta t[/tex]): (\Delta v /\Delta t[/tex]).
 
I got it now. Thank you da_willem
 

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