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

MATH100 December 2011

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[hide] 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 $s=f(t)$ .

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[show] Hint
The acceleration is the rate of change of velocity.

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[show] Solution
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: $\displaystyle f(t)=2t^{3}-15t^{2}+36t$