Science:Math Exam Resources/Courses/MATH110/December 2014/Question 05 (b)

MATH110 December 2014

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Question 05 (b)
If a rock is thrown upward on the planet Mars with a velocity of 10 m/s, its height (in metres) above the ground level after ${\textstyle t}$ seconds is given by $$h(t)=10t-2t^{2}.$$ (b) Find the velocity of the rock when it hits the surface.

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
The velocity of the rock means the rate of change of the height with respect to the change of the time. In other words, it is given by ${\frac {dh}{dt}}=h'(t)$.

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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. As we mentioned in the Hint, the velocity $v(t)$ of the rock is given by $v(t)=h'(t)$. Since the rock hits the surface at $t_{h}=5$(s), what we need to find is $v(5)$(m/s). First, using the power rule, we get the derivative: $v(t)=h'(t)=(10t-2t^{2})'=10-4t$. Then, plugging $t_{h}=5$, we get $\color {blue}v(5)=10-20=-10m/s$