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Mth 173
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Book Problem 6.3.19
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A car starts from rest at time t=0 and accelereates at -0.6t+4 m/sec^2 for 0 <= t <= 12. How long does it take for the car to go 100 meters.
I can't figure it out. Acceleration is the derivative of velocity which in this case is -0.3t^2 + 4t. Velocity is the derivative of distance which in this case is -0.1t^3 + 2t^2 + c.
I know you make -0.1t^3 + 2t^2 + c = 100 and solve for t. I know I probably should, but I don't remember how to solve this. I have wasted too much time trying to figure this out on my own. I wanted to on my own because I should know how, I'm afraid I need some help.
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Don't hesitate to ask for help when you get bogged down on a problem. You can judge the point at which your efforts are yielding diminishing returns.
The change in position between two clock times is the definite integral of the velocity function between those times. So you would set the definite integral equal to 100:
integral( -0.3 tau^2 + 4 tau, tau from 0 to t) = 100
We use tau as the variable of integration because we want to use t as one on the limits on our integral, and it would be confusing to use t within the integral as well as for one of the limits. We could equally well have used any other symbol for the variable of integration, but tau is a fairly natural selection here.
You will end up with the equation -0.1 t^3 + 2 t^2 = 100.
You could equally well have used your function x(t) = -0.1 t^3 + 2 t^2 + c. With this function
the position at t = 0 is x(0) = c
the position at clock time t is -0.1 t^3 + 2 t^2 + c
So the change in position is
`dx = x(t) - x(0) = -0.1 t^3 + 2 t^2 + c - c = -0.1 t^3 + 2 t^2.
Set the change in position equal to 100. This will give you the same equation as before.
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