# [SOLVED] Basic Integration

#### unstopabl3

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:

#### drumist

Yes, it looks correct.

unstopabl3

#### unstopabl3

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
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.

unstopabl3

#### unstopabl3

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!

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