# Math Help - Finding distance, given velocity with cos.

1. ## Finding distance, given velocity with cos.

What a wonderful website this is. Thank you all in advance for any help or guidance you can give me.

So, okay.

Find the distance traveled in 15 seconds by an object moving at a velocity of v(t) = 20 + 7 cos t feet per second.

Again, thank you for any help. I have no idea what to do with the cosine.

Edit: You know, I put this here because I was told it had to do with integrals? It might fit better in Trigonometry though. Sorry about that!

2. Originally Posted by AppleTini
What a wonderful website this is. Thank you all in advance for any help or guidance you can give me.

So, okay.

Find the distance traveled in 15 seconds by an object moving at a velocity of v(t) = 20 + 7 cos t feet per second.

Again, thank you for any help. I have no idea what to do with the cosine.

$s(t)=\int v(t)\,dt$

Therefore, $s(t)=\int (20+7\cos t)\,dt=20t+7\sin(t){\color{red}+C}$

Is there an initial condition for position?

--Chris

3. Originally Posted by Chris L T521
Is there an initial condition for position?
No, it doesn't seem so.

The problem is that I'm taking this AP Calculus class online [bad idea, for the record], and the online book really is not helpful at all. It hasn't actually talked about integrals or anything, and I certainly don't really remember them from two years ago. So I'm pretty much on my own. Thank you so much, though.

4. It MAY seem logical to assume one means the first 15 seconds.

The integral on [0,15] gives 300 + 7*sin(15) = 304.552

Unfortunately, it may NOT be a logical assumption. If we don't start watching for five seconds, the integral on [5,20] gives 300 + 7(sin(20)-sin(5)) = 313.103

We could play this game for a long time. One needs those initial conditions or some other guidance as to when or where we are clocking the 15 seconds.

In any case, you can estimate the answer well enough.

The constant "20" is, well, constant. It ALWAYS manages 300 in 15 seconds. The tricky part is the cosine, as you said. Interestingly, the cosine is periodic. Every $2\pi$ it has gotten you nowhere. In this way, we can eliminate some of the cosine problem. $15 = 2*(2\pi) + 0.387*(2\pi)$. This means you really have to worry only about the last 2.434 seconds of the trip, since the previous $4\pi$ seconds didn't get us anywhere! Of course, you still have to know where to start or stop. Just any 15 seconds will not due. We need to know which ones to watch.

5. Originally Posted by AppleTini
What a wonderful website this is. Thank you all in advance for any help or guidance you can give me.

So, okay.

Find the distance traveled in 15 seconds by an object moving at a velocity of v(t) = 20 + 7 cos t feet per second.

Again, thank you for any help. I have no idea what to do with the cosine.

Edit: You know, I put this here because I was told it had to do with integrals? It might fit better in Trigonometry though. Sorry about that!
Actually the distance s(t) is given by
$s(t) = \int |v(t)|~dt$

Fortunately in this case you get the same answer as has been given in the previous posts.

-Dan

6. displacement (a vector quantity) = $\int_{t_1}^{t_2} v(t) dt$

distance traveled (a scalar quantity) = $\int_{t_1}^{t_2} |v(t)| dt$

7. Well to find distance you have to integrate the whole v(t) funcn.

integrating dat you will get
s(t) = 20t + 7sint

put t=15
you will get

s= 140 + 7sin15

sin15 =.26 approx.
so you get the ans..

8. Originally Posted by vishalgarg
Well to find distance you have to integrate the whole v(t) funcn.

integrating dat you will get
s(t) = 20t + 7sint

put t=15
you will get

s= 140 + 7sin15

sin15 =.26 approx.
so you get the ans..
How does 20 times 15 equal 140?

-Dan