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

MATH102 December 2011

  •  Q1 (a) i  •  Q1 (a) ii  •  Q1 (b) i  •  Q1 (b) ii  •  Q1 (b) iii  •  Q1 (c)  •  Q2 (a) i  •  Q2 (a) ii  •  Q2 (a) iii  •  Q2 (b) i  •  Q2 (b) ii  •  Q2 (b) iii  •  Q2 (c)  •  Q3  •  Q4 (i)  •  Q4 (ii)  •  Q4 (iii)  •  Q4 (iv)  •  Q4 (v)  •  Q4 (vi)  •  Q5  •  Q6  •  Q7 (i)  •  Q7 (ii)  •  Q7 (iii)  •  Q8 (i)  •  Q8 (ii)  •  Q8 (iii)  •  Q9  •  

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Question 08 (iii)

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. (iii) A few days later you are shopping for a new car. You would like to purchase a faster car to avoid moose accidents. What is the minimum acceleration a that your car would have needed to avoid being caught by the moose in the previous scenario?

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
Follow the approach from part (i), but replace the acceleration 2m/s with the variable a. Then choose a such that there is no solution to the equation for the hitting time from (ii)

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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. We consider the same problem as in part (i), but now we have a car with acceleration a. We will set up the equation for the distance between the moose and car and try to determine values of a such that there is no time where the distance between the car and the moose is equal to 0. As before, the position of the moose as a function of time, m(t), is given by: $\displaystyle m(t)=12-8t.$ The position of the car however is now given by $\displaystyle c(t)=-{\frac {a}{2}}t^{2}.$ Hence the distance between the moose and the car is given by $d(t)=|c(t)-m(t)|=\left|-{\frac {a}{2}}t^{2}+8t-12\right|.$ What we want to do now is to choose a value for the acceleration of the car, a, such that the equation d(T) = 0 has no (real) solution for T. (i.e: Choose a so that there is no time where the distance between car and moose is 0). $d(T)=\left|-{\frac {a}{2}}T^{2}+8T-12\right|=0\quad \rightarrow \quad T={\frac {-8\pm {\sqrt {64-24a}}}{-a}}$ To ensure that there are no real solutions, we must look for values of a such that 64 - 24a is less than zero. $64-24a<0\quad \rightarrow \quad {\color {blue}a>{\frac {8}{3}}}$ Therefore, we need a car that has acceleration greater than 8/3 to ensure that the moose will not hit our car next time this exact situation occurs.

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