Science:Math Exam Resources/Courses/MATH100/December 2011/Question 03 (a)
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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 .
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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: