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!
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?
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 ...
Better check for typos. Also, answer keys have been known to be wrong ...
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.