Science:Math Exam Resources/Courses/MATH103/April 2009/Question 01 (e)

MATH103 April 2009

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

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Question 01 (e)

Multiple Choice question: Select ONE correct answer. You will not be graded for any work. A charged molecule moves in a changing electric field which results in an acceleration of a(t) = cos(t) (in appropriate units). At time t = 0, the molecule is at rest at position x(0) = 1. What is the position x(t) when the molecule's velocity reaches the value -1 for the first time after time 0? (a) -1 (b) 0 (c) 1 (d) 2 (e) The velocity never reaches -1

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
The acceleration $a(t)\$ is the second derivative $\ a(t)=x''(t)$ of the position $\ x(t)$.

Hint 2
With $\ a(t)=x''(t)$ we find, that we need to solve the differential equation ${\begin{aligned}x''(t)&=\cos(t),\\x'(0)&=0\quad {\text{(the molecule rests at time 0)}}\\x(0)&=1\end{aligned}}$

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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 find, that we need to solve the differential equation ${\begin{aligned}x''(t)&=\cos(t)\\x'(0)&=0\\x(0)&=1\end{aligned}}$ We integrate $\ x''(t)=\cos(t)$ and get $x'(t)=\sin(t)+c_{1}\$. With $0=x'(0)=\sin(0)+c_{1}=c_{1}\$ we find $c_{1}=0\$. We continue with integrating $x'(t)=\sin(t)\$, which leads to $\ x(t)=-\cos(t)+c_{2}$. With $\ x(0)=1$ we find $\ 1=x(0)=-\cos(0)+c_{2}=-1+c_{2}$ and get $\ c_{2}=2$. The velocity beeing $\ x'(t)=\sin(t)$ reaches $\ -1$ at $t={\frac {3}{2}}\pi \$, and we get $\ x\left({\frac {3}{2}}\pi \right)=-\cos \left({\frac {3}{2}}\pi \right)+2=-0+2=2$. So, the answer is (d).

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