Science:Math Exam Resources/Courses/MATH100/December 2011/Question 03 (b)
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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.
The velocity after 2 seconds is 0, and ƒ(0) = 0.
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!
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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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
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 and so, setting this equal to zero and solving gives
and hence the critical points are at . Since we only care about time up to , we only need to check whether the point at is a maximum or minimum. Using the second derivative test we see that
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:
The total distance traveled from t=0 to t=2 is:
Then, we can calculate the total distance traveled from t=2 to t=3:
We add these two values together to obtain the total distance traveled in the first three seconds, which we find to be 29 metres.