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

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 $s=f(t)$, $t\geq 0$, where $t$ is measured in seconds and $s$ in meters. The acceleration function $a(t)$ is given by $\displaystyle a(t)=12t-30$ The velocity after 2 seconds is 0, and $f(0)=0$. (a) Find the position function $s=f(t)$.

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!

Hint
The acceleration is the rate of change of velocity.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $\displaystyle f''(t)=a(t)$ and so $\displaystyle f''(t)=12t-30$ Since this is a polynomial in t of degree one, we know that the velocity must be a polynomial of degree 2: $\displaystyle f'(t)=at^{2}+bt+c$ whose derivative is 12t-30 ${\frac {d}{dt}}\left(at^{2}+bt+c\right)=2at+b$ Hence we have $\displaystyle 2a=12\qquad {\text{and}}\qquad b=-30$ And so the velocity is $\displaystyle f'(t)=6t^{2}-30t+c$ for some constant c. Since the velocity after 2 seconds is 0, we actually are being told that ƒ'(2)=0 and hence $\displaystyle 6\cdot 4-30\cdot 2+c=0\qquad \Rightarrow c=36$ And so the velocity is $\displaystyle f'(t)=6t^{2}-30t+36$ 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 $\displaystyle f(t)=At^{3}+Bt^{2}+Ct+D$ and computing its derivative gives us $\displaystyle f'(t)=3At^{2}+2Bt+C$ Hence $\displaystyle 3A=6\qquad {\text{and}}\qquad 2B=-30\qquad {\text{and}}\qquad C=36$ So the position function is $\displaystyle f(t)=2t^{3}-15t^{2}+36t+D$ and we can find out the value of D since we are told that ƒ(0) = 0 which gives $\displaystyle D=0$ And so finally, we obtain that the position function is: $\displaystyle f(t)=2t^{3}-15t^{2}+36t$

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