[SOLVED] Basic Integration

Sep 2009
148
2
This is a mechanics question but since it requires calculus to solve it, I am posting it here.

\(\displaystyle v=k(200t-t^2)\)

Where

Distance= 200m
Time= 100s

My Attempt

\(\displaystyle ds/dt=k(200t-t^2)\)
\(\displaystyle \int{ds}=\int{k(200t-t^2)dt}\)
\(\displaystyle s=k(\frac{200t^2}{2}-\frac{t^3}{3})+c\)

Have I done the above 3 steps correctly?
 
Last edited:
Sep 2009
148
2
So if I go on to solve the whole equation...

My Attempt

\(\displaystyle ds/dt=k(200t-t^2)\)
\(\displaystyle \int{ds}=\int{k(200t-t^2)dt}\)
\(\displaystyle s=k(\frac{200t^2}{2}-\frac{t^3}{3})+c\)

When t=0 then s=0 and c=0

Thus, c=0

Our equation becomes...

\(\displaystyle s=k(\frac{200t^2}{2}-\frac{t^3}{3})\)

Substituting s=200 when t=100

\(\displaystyle 200=k(\frac{200(100)^2}{2}-\frac{(100)^3}{3})\)

\(\displaystyle 200=k(\frac{200(10000)}{2}-\frac{(1000000)}{3})\)

\(\displaystyle 200=k(1000000-333333.3333)\)

\(\displaystyle 200=k(666666.6667)\)

\(\displaystyle \frac{200}{666666.6667}=k\)

\(\displaystyle k=0.0003\)

All of my steps are correct?
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Yes, although I can't help wondering why you keep writing "\(\displaystyle \frac{200}{2}\) rather than just "100". Also, while you have solved for k, you haven't yet written the solution itself.
 
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Sep 2009
148
2
Well that's because I wanted to show most of my steps ;)

I have already solved the question, just wanted to make sure that all of my steps/working was correct!

Thanks for the quick response!