What is the displacement of the particle

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Here's my problem, The velocity of a particle is given by
v = (5.5 m/s^4)t^3 - (5.4 m/s), what is the displacement of the particle during the interval t=3.5s to t=9.0s? I don't even know if I started it off right. I recalled seeing something similar to this in calc, so I tried integrating the formula given using 3.5 as 'a' and 9.0 as 'b'. I ended up getting 80440.5 as my answer, but it seems a bit large. I just need to know how to approach the problem. Thanks
 
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HallsofIvy

Science Advisor
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Re: Displacement

"a" and "b"? I don't understand what those are!! Oh, you mean the limits of integration. That's exactly what you should do.

What you need to know is that velocity is the derivative of position so position is the anti-derivative of velocity- that is, the integral.

Since v= dx/dt= (5.5)t^3 - 5.4, x= (5.5/4)t^4- 5.4t+ x0. (x0 is the "unknown constant" and would be the position of the particle at t=0). Evaluate at t= 9.0s and t= 3.9s and subtract. (The x0 in each term will, of course, cancel).

Why does your answer seem "a bit large"? With the t^3 term the speed is going to increase rapidly!
 
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Hey thanks for the help. I figured out what I did wrong. I integrated incorrrectly. THANKS!!
 

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