Science:Math Exam Resources/Courses/MATH102/December 2011/Question 08 (i)

MATH102 December 2011

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[hide] Question 08 (i)
You are driving down the highway when you see a sleeping moose. You apply the brakes and carefully stop your car 20m away from the animal. While you are looking for your camera the moose wakes up. It instantly charges toward your car at a constant speed of 8m/s. One second later, you start backing away from the moose at a constant acceleration of 2m/s². (i) Write down a function d(t) that is the distance from your car to the moose where t = 0 indicates the moment when you start backing away.

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[show] Hint
First, consider the displacement of the car and the moose separately. For example, fix the initial position to be where the car is stopped. Relatively to that point, what are the respective distance of the car and the moose over time. Then, you'll find the distance between the car and the moose by taking the difference of those two things.

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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. Let $d(t)$ be the distance between your car and the moose, where t = 0 represents the time when your car begins backing up. The moose began charging one second before t=0, i.e. at t = -1, and so, when the car begins backing up the moose is 20m-8m/s(1s) = 12m away. Let the origin be the position of the car at time t = 0. Consider the moose. If the moose is 12m away at t = 0* and is moving towards the car at a speed of 8 m/s, we can define the position of the moose as m(t) where $\displaystyle m(t)=12-8t.$ If we set $c(t)$ be the position of the car at time t. The car is accelerating away from the moose at 2 m/s², then $\displaystyle c''(t)=-2\quad \rightarrow \quad c'(t)=-2t+c_{1}$ Since the car is not moving at t = 0, we have that $c'(0)=c_{1}=0$ . Thus $c'(t)=-2t$ . From this, we obtain that $c(t)=-t^{2}+c_{2}$ . Since we set the position of the car at t = 0 to be $c(0)=0$ , then $c(t)=-t^{2}$ . The distance between the car and the moose is thus (note that the distance is always a positive number) $d(t)=\|c(t)-m(t)\|={\color {blue}\|-t^{2}+8t-12\|}$