Asked by Kyle

Find the distance, in feet, a particle travels in its first 2 seconds of travel, if it moves according to the velocity equation v(t)= 6t2 − 18t + 12 (in feet/sec).

A. 4

B. 5

C. 6

D. -1

Please helpp, I'm struggling!
Thank you in advance for your help!

Answers

Answered by oobleck
v(t) = ds/dt
so, the distance s is
s(t) = ∫ v(t) dt = 2t^3 - 9t^2 + 12t

So, what do you think?
Answered by Justin
So I guess now that we have the integral, I substitute the t=2 into the equation, right?
Answered by oobleck
good guess.
Answered by Justin
I got and selected 4 after substituting t=2, but when I selected that one I got marked wrong...
Answered by Ms. Sue
Kyle/Justin -- please use the same name for your posts. I almost deleted your Justin posts as someone horning in on another's question.

Answered by oobleck
I get 4 as well.
Better check for typos. Also, answer keys have been known to be wrong ...
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