MATH100 December 2011
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Question 03 (a)
Full-Solution Problems. Justify your answers and show all your work. Simplification of answers is not required unless explicitly stated.
A particle moves in a straight line according to a law of motion , , where is measured in seconds and in meters. The acceleration function is given by
The velocity after 2 seconds is 0, and .
(a) Find the position function .
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!
The acceleration is the rate of change of velocity.
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If s = ƒ(t ) is the function describing the position of the particle, then its derivative ƒ'(t ) is its velocity and its second derivative the acceleration. This means we know that
Since this is a polynomial in t of degree one, we know that the velocity must be a polynomial of degree 2:
whose derivative is 12t-30
Hence we have
And so the velocity is
for some constant c. Since the velocity after 2 seconds is 0, we actually are being told that ƒ'(2)=0 and hence
And so the velocity is
Now, the velocity being the derivative of the position function, we can deduce that it has to be a polynomial of degree 3 and write
and computing its derivative gives us
So the position function is
and we can find out the value of D since we are told that ƒ(0) = 0 which gives
And so finally, we obtain that the position function is:
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