Question 5
5 pts 1 Details
A particle moves along a straight line and its position at time $t$ is given by $s(t) = 2t^3 - 21t^2 + 60t$ where $s$ is
measured in feet and $t$ in seconds.
Find the velocity (in ft/sec) of the particle at time $t = 0$:
ft/sec
The particle stops moving (i.e. is in a rest) twice,
first when $t =$
seconds
and again when $t =$
seconds
What is the position of the particle at time 14?
ft
Finally, what is the TOTAL distance the particle travels between time 0 and time 14?
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Question 6
5 pts 1 Details
This problem, showing all your numerical work, must be submitted after you complete the Exam to get
full credit.
Let $y = \ln\left(\frac{x^2 - 6x + 5}{x - 5}\right)$
$y =$
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