Dynamics problem

  • #1
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this is a part of a much bigger problem, I have a 600lb motorcycle going from point A to point B, which are 500 ft apart, i know the force exerted by the road on the motorcycle when the throttle is pressed is F(t)=300(1-.1t).
I need time in order velocity at B ,
I'm really lost on where to start on this, i was thinking f=ma could give me acceleration, but then i dont know what to do with that.
 

Answers and Replies

  • #2
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Are you trying to determine the velocity at point B?

If you know the acceleration then by definition you should be able to get the velocity.
 
  • #3
so f=ma will give me a acceleration as a function of t, and yes i need velocity at B, but then won't i only have velocity as a function of t if by definition?.
 
  • #4
RUber
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Right, but if you have velocity, you can find position. Find how long it takes to get to point B, then use that t for your velocity.
 
  • #5
SteamKing
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this is a part of a much bigger problem, I have a 600lb motorcycle going from point A to point B, which are 500 ft apart, i know the force exerted by the road on the motorcycle when the throttle is pressed is F(t)=300(1-.1t).
I need time in order velocity at B ,
I'm really lost on where to start on this, i was thinking f=ma could give me acceleration, but then i dont know what to do with that.

Gee, if only there were a way that acceleration could be used to get velocity, and that velocity could then be used to find distance. It would seem to be a good reason to study rectilinear motion in physics, or something. Maybe somebody wrote a textbook on how to do this stuff. IDK, that would probably take a lot of work to figure out.
 
  • #6
so would i use Vi^2=Vf^2+2A(500) and substitute in V and A as their function of t equations? because then i get t= 10.05 does that seem right?
 
  • #7
RUber
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What units are you using for m, F, and distance? I hope F is in ft/lbs.
 
  • #9
haruspex
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so would i use Vi^2=Vf^2+2A(500) and substitute in V and A as their function of t equations? because then i get t= 10.05 does that seem right?
No, that formula only works for constant acceleration. You need to use the general relationships between acceleration, velocity, and distance - namely, integral formulas.
 

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