Using Calculus to Solve for Time in a Particle's Position Function

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The discussion revolves around finding the time at which a particle, described by the position function s = t^3 - 4.5t^2 - 7t, reaches an instantaneous velocity of 5 m/sec. Participants emphasize the need to calculate the derivative of the position function to derive the velocity function. The derivative is obtained by expanding the terms and simplifying, leading to the expression s'(t) = 3t^2 - 9t - 7. Once the derivative is found, it can be set equal to 5 to solve for the corresponding time values. The conversation highlights the importance of understanding derivatives in relation to position and velocity functions.
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Homework Statement



(Sorry I don't know how to insert nice looking equations)

If the position function of a particle is given by
s = t3 - 4.5t2 - 7t, t >= 0
, the particle reaches an instantaneous
velocity of 5 m/sec when t =
1
2
3
4
5


Homework Equations




The Attempt at a Solution


Well, the problem I'm having is how to do this to find t instead of finding the velocity. If I attempt to find the derivative, I always get stuck at something like:

lim h -> t [(t+h)3 - 4.5(t+h)2 - 7(t+h)] - [t3 - 4.5t2 - 7t] ALL OVER h

I can play with that a little bit, but it never gets anywhere. thanks
 
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Assuming that you have to use the definition of the derivative, expand your (t + h)3 and (t + h)2 terms and then combine all like terms in the numerator. There should be some simplification so that you can then divide by the h in the denominator. Finally, take the limit as h --> 0 and you will be left with your derivative.
 
Well you need to find the derivative. I can tell you the answer is 4 but you should find it for yourself. If that is where you get stuck, expand it out and cancel out like terms. What i mean by that is calculate (t+h)(t+h)(t+h) and then find 4.5(t+h)(t+h) and so on. You will see that all terms without the h cancel out, and you are able to divide everything by h.

Keep in mind that the answer you will then have is now a speed function, not a position function. If you are still stuck post again.
 
OK, I did that, and assuming my algebra was right I got 3t2 - 9t - 7

And here I'm stuck again. Do I just set that equal to 5 and solve?
 
That is the correct value of s'(t). Now evaluate s'(t) at t = 1, 2, 3, 4, and 5.
 
Awesome now I see. Thanks for the help Mark and dacruick.

Ill probably have a few more questions over the next few hours.
 
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