Find the distance traveled in 15 sec by an object moving with a velocity of v(t)=20+7cost ft/sec. Do I use calculus? And how do I solve this problem? This is my first calculus class, so please help.

Question

aravindg · Answer

distance travelled=velcity $\times$ time

anonymous · Answer

So would I do d=15sec(20+7cost ft/sec)? and t by cos means 15sec?

anonymous · Answer

You would take the integral of the velocity from 0 to 15, which would give you the total displacement of the object, which is different from distance.  However, your velocity is always positive in this case, so your displacement equals your distance.

So if

$$v = 20 + 7\cos t$$

Then your integral is

$$\int\limits_{0}^{15}v = \int\limits_{0}^{15}20 + 7cost$$

We can split these integrals, which will make things easier.  This gives us:

$$\int\limits_{0}^{15} 20 +7 \int\limits_{0}^{15}cost$$

Integrating these gives us:

(20*5 - 0*5) + 7(sin(15) - sin(0))

Simplifying this gives us

100 + 7sin(15)

anonymous · Answer

You can't just do velocity x time in this case because your velocity varies with time.

aravindg · Answer

we could use average velocity $\times $ time

anonymous · Answer

thanks!

anonymous · Answer

however, where does the 5 come from in (20*5 - 0*5) + 7(sin(15) - sin(0))?

anonymous · Answer

It should be 15, not 5, sorry about that.

anonymous · Answer

It's ok. So it should be (20*15 - 0*15) + 7(sin(15) - sin(0))?

or (20*15 - 0*0) + 7(sin(15) - sin(0))?

and why must sin be used instead of cos?

anonymous · Answer

it should be

(20*15 - 20*0) + 7*sin(15)

Sin gets used instead of cos because the anti-derivative of cos is sin.

To find a definite integral, you take the antiderivative of what is inside the integral, which is cos(t) in this case, and then apply the bounds to it.

anonymous · Answer

Thank you very much! You are very helpful