Science:Math Exam Resources/Courses/MATH110/December 2016/Question 06 (b)

MATH110 December 2016

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Question 06 (b)

Explorers on a small airless planet used a spring gun to launch a small ball vertically upward from the ground at a launch velocity of $15{\text{m}}/{\text{s}}$. The explorers expected the ball to reach a height of $$s(t)=15t-(1/2)gt^{2}$$ metres at $t$ seconds after being launched, where $g$ is a constant. (b) Sketch the graph of the ball’s velocity $v(t)$ for $0\leq t\leq 40$. Make sure you indicate the values of any intercepts, if they exist.

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
Find the derivative of the height function given in part (a), which is the velocity $v(t)$.

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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. Using the constant $g={\frac {3}{4}}$ found in part (a), the height of the ball is given by $$s(t)=15t-{\frac {3}{8}}t^{2}$$ As suggested in the hint, the velocity is a derivative of the height function; $$v(t)=s'(t)=\left(15t-{\frac {3}{8}}t^{2}\right)'=15-{\frac {3}{4}}t$$ . By plugging $t=0$, we get a $y$-intercept as $y=15$. On the other hand, solving $v(t)=15-{\frac {3}{4}}t=0$, we have a $t$-intercept $t=20$. Since its graph is a straight line with slope $-{\frac {3}{4}}$ passing through the intercepts $(0,15)$ and $(20,0)$, we can sketch it as follows.