Science:Math Exam Resources/Courses/MATH110/April 2018/Question 05 (a)

MATH110 April 2018

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q2 (e)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q4 (d)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q6 (a)  •  Q6 (b)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q7 (d)  •  Q7 (e)  •  Q8 (a)  •  Q8 (b)  •  Q8 (c)  •  Q8 (d)  •  Q9  •  Q10  •  

   Other MATH110 Exams

  •  December 2011  •  December 2010  •  April 2012  •  April 2011  •  December 2012  •  April 2013  •  December 2013  •  April 2014  •  December 2014  •  December 2015  •  April 2017  •  April 2016  •  December 2016  •  December 2017  •  April 2018  •  

Question 05 (a)
A stone is shot vertically into the air at an initial velocity of 16 m/s. Its height $s$ (in metres) above the ground after $t$ seconds is given by $$s(t)=16t-5t^{2}.$$ (a) For how many seconds does the stone rise before starting to fall back down?

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
We expect that the stone rises and then falls. So, find time at which $s$ has its maximum value.

Hint 2
An extreme value of $s$ can only occur at a critical number.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. By Hint 1, we find $t_{0}>0$ at which the height function $s$ has its maximum value. Then, since $t_{0}$ is a local maximum, it is a critical number. Note that $s$ is a polynomial and hence differentiable on $(0,\infty )$. Also, $t_{0}$ is not an endpoint. Therefore, by Hint 2, we have $$s'(t_{0})=0.$$ By the power rule for derivatives, we get $$s'(t)=(16t-5t^{2})'=16-10t.$$ Solving $s'(t_{0})=16-10t_{0}=0$, we obtain $t_{0}={\frac {16}{10}}={\frac {8}{5}}.$ Considering that for $0<t<{\frac {8}{5}}$, $s'(t)=10\left({\frac {8}{5}}-t\right)>0$, the height $s$ is increasing on $\left(0,{\frac {8}{5}}\right)$. On the other hand, for ${\frac {8}{5}}<t<\infty$, $s'(t)=10\left({\frac {8}{5}}-t\right)<0$, so that the height $s$ is decreasing on $\left({\frac {8}{5}},\infty \right)$. Therefore, the desired time $t_{0}$ is ${\frac {8}{5}}$. Answer: $\color {blue}{\frac {8}{5}}$.