HELP ASAP!!!!!!!!
The position of a particle is given by the equation below, where s is in feet and
t is in seconds. How long does it take for the velocity to be equal to zero?

Question

HELP ASAP!!!!!!!!
The position of a particle is given by the equation below, where s is in feet and 
t is in seconds. How long does it take for the velocity to be equal to zero?

adunb8 · Answer

$$s= 3t^3+50t^2+150t - 20$$

adunb8 · Answer

a. t=4.66s and 1.79 s
b. t=9.32 s and 1.79 s
c. t=8.32 s and .788 s
d. None of these.
e. t=4.32 s and 1.79 s

matt101 · Answer

The velocity is the rate of change of the position. This means we need to find the derivative of the position equation to come up with the velocity equation:

$$\frac{ ds(t) }{ dx } =v(t)= 9t^2+100t+150$$

We're trying to find the time when velocity is 0, so just solve for t using whatever method you like after setting v(t)=0:

$$0=9t^2+100t+150$$$$t=-1.79,-9.32$$
Both times are negative; this means the particle had 0 velocity BEFORE we started timing, but from the time we start tracking the particle's position, the velocity will NEVER be equal to 0. You can see from the velocity equation that no matter what value you have for t > 0, you always end up adding a bunch of positive numbers and so your velocity is always going to be positive.

Your answer is D.

adunb8 · Answer

answer is actually B. i think you might of messed the signs but this is great info. Thank you!

matt101 · Answer

Glad I could help! Just double checked my numbers, and for the position equation you gave me, the times I got are the solutions. Maybe there's a mistake in the original equation?