Question

Find the distance traveled by an object on the interval [0, 5] if its velocity is v(t)= sin((pi/3)t)

Answer 1

Area under v-t graph gives you displacement. Integrate the equation from t = 0 to t = 5 to get the total distance traveled.

Answer 2

I think "distance traveled" means: how far is the object from the starting point after 5 seconds. In that case you can ignore the to and fro movements caused by the sine function. So you do exactly as @navk said: just calculate \(\int_0^5 \sin(\frac{\pi}{3}t)dt\).
Do you know how to do this?

Answer 3

i tryed that but the only possible answers are

\[21/2\pi\]

\[3/2\pi\]

7/2

1/2

Answer 4

wait ill try it again

Answer 5

Well, if you calculate this integral, you will get one of the 4 answers!

Answer 6

@bestud: please share your actual work; that would make it much easier for others to give you feedback on what you have done.

Answer 7

5sin(πt3) = 5sin(1.047197551196598t)

this is what i got

Answer 8

i dont really understand integrals

Answer 9

Do you know a primitive function of \(\sin(\frac{\pi}{3}t)\)?

Answer 10

Did you integrate it? First integrate sin(pi/3)t and then evaluate it at the limits 0 to 5

Answer 11

That is why you are on OpenStudy!
You have to find a function whose derivative is \(\sin\frac{\pi}{3}t\)?

Answer 12

Knowing the derivatives of sin and cos would help...I'm sure you know these!

Answer 13

Also, the Fundamental theorem of calculus is needed:

\(\int_a^bf(x)dx=F(b)-F(a)\), where F is a primitive function of f: this means F'=f.

That is why you have to find a primitive function of \(\sin\frac{\pi}{3}t\).

Answer 14

Take a look at the Fundamental Theorem: it makes everything so much easier!
That integral, which is essentially an infinite sum of infinitely small numbers (how could one ever hope to find out what it adds up to!) can be calculated byjust subtracting two values: F(b) and F(a).

So the only thing you have to do is: find F.
So find a function whose derivative is \(\sin\frac{\pi}{3}t\).

Answer 15

You know the derivatives of sin and cos:

\((\sin t)'=\cos t\)
\((\cos t)'=-\sin t\)

Answer 16

So if you were looking for a function whose derivative is \(\sin t\), you would come up with \(-\cos t\). The extra minus sign is needed to give sin after differentiation.

Answer 17

But what is the primitive of \(\sin\frac{\pi}{3}t\)?
If we try this: \(-\cos\frac{\pi}{3}t\), we get \(\sin\frac{\pi}{3}t\cdot \frac{\pi}{3}\) after differentiation.
Now there is an extra factor \(\frac{\pi}{3}\) that we don't want.

So take as primitive the function \(F(t)=-\frac{3}{\pi}\cos\frac{\pi}{3}t\).

Answer 18

If you calculate F'(t), you will see that the \(\frac{3}{\pi}\) cancels out the unwanted \(\frac{\pi}{3}\), so everything is fine.

Now \(F(b)-F(a)=F(5)-F(0)=-\frac{3}{\pi}\cos\frac{5\pi}{3}--\frac{3}{\pi}\cos0\).

I think if you evaluate this, you will be able to see which of your answers is right.

Answer 19

BTW, you may not be familiar with the term "primitive function".
It is also called the "antiderivative".

Answer 20

yep i got it : 3/2pi !!!! looking at antiderivatives is much easier thanks !!!!

Answer 21

YW!