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January 31, 2015

January 31, 2015

Posted by **Anonymous** on Thursday, May 22, 2008 at 2:07am.

v(t)=25tsin(t^2)

At time=0, you were backed up against the bumper of your car.

How far were you from your car two minutes after you started walking?

Integral (0 to 2) v(t) dt = 20.671

What was your acceleration 1 minute into your walk?

V’(1)=48.052

At what time was your instantaneous velocity as your average velocity over the first minute? (how do I solve this?)

- Calculus -
**drwls**, Thursday, May 22, 2008 at 5:30amYour V'(1) looks OK.

It is 50 cos 1 + 25 sin 1

I didn't check your integral.

Your wording of the last question, with the word "as", doesn't make sense to me. Do you mean to ask

"At what time was your instantaneous velocity equal to your average velocity over the first minute?"

To answer this, set

v(t) = (distance traveled)/t = x(t)/t

Use your x(t) solution to get that.

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