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

MATH100 December 2011

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[hide] Question 03 (b)
Full-Solution Problems. In questions 2-8, 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 = ƒ(t ), t ≥ 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 ƒ(0) = 0. Find the total distance travelled by the particle during the first three seconds.

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[show] Hint
When is the position function increasing? When is it decreasing? Remember that in calculating the total distance means that travelling backwards is still covering distance, therefore adding (not subtracting) to the total distance travelled.

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[show] 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. We are given acceleration and asked to argue about distance. Since the derivative of a position function is velocity and the derivative of a velocity function is acceleration, we with to antidifferentiate our original function twice. This was done in part (a) giving the answer of $\displaystyle f(t)=2t^{3}-15t^{2}+36t$ As this is a position function, we need to check when the particle reverses direction. This occurs potentially at critical points. We know the derivative of ƒ(t) is $v(t)$ and so, setting this equal to zero and solving gives $\displaystyle 0=f'(t)=6t^{2}-30t+36=6(t^{2}-5t+6)=6(t-3)(t-2)$ and hence the critical points are at $t=2,3$ . Since we only care about time up to $t=3$ , we only need to check whether the point at $t=2$ is a maximum or minimum. Using the second derivative test we see that $\displaystyle f''(2)=v'(2)=a(2)=12(2)-30=-6<0$ and thus t=2 is a maximum of ƒ. In particular, this means that our particle changes directions at t=2. Thus, we calculate the total distance traveled up to t=2 first then from t=2 until t=3: $\displaystyle {\begin{aligned}f(0)&=2(0)^{3}-15(0)^{2}+36(0)\\&=0\end{aligned}}$ $\displaystyle {\begin{aligned}f(2)&=2(2)^{3}-15(2)^{2}+36(2)\\&=2(8)-15(4)+72\\&=16-60+72\\&=28\end{aligned}}$ $\displaystyle {\begin{aligned}f(3)&=2(3)^{3}-15(3)^{2}+36(3)\\&=2(27)-15(9)+108\\&=54-135+108\\&=27\end{aligned}}$ The total distance traveled from t=0 to t=2 is: $\displaystyle {\begin{aligned}\|f(2)-f(0)\|&=\|28-0\|&=28\end{aligned}}$ Then, we can calculate the total distance traveled from t=2 to t=3: $\displaystyle {\begin{aligned}\|f(3)-f(2)\|&=\|27-28\|&=1\end{aligned}}$ We add these two values together to obtain the total distance traveled in the first three seconds, which we find to be 29 metres.