Question

A particle moves on x-axis blah blah... its initial position is x(0)=27 and velocity is v(t)=4t^3-30t^2+72t-54 on the time interval 0
a) write an expression for the position function (general)
t^4-10t^3+36t^2-54t+C
b) when does the particle change direction?
help
c) find t when particle is moving to the right
help
d) find velocity when acceleration is 0.
ez, don't need help for this one.

Answer 1

b) The particle changes direction when its velocity changes sign. Use the first derivative test to find the critical points:

$v(t)=2(t-3)^2(2t-3)=0\implies t=\dfrac32\,\text{ or }t=3$

Then for $t$, we have $v(t)$; for $\dfrac32$, we have $v(t)>0$; and for $t>3$, we have $v(t)>0$ again. These three facts indicate that the velocity only changes once at $t=\dfrac32$.

c) Pick "to the right" to be the positive direction. Then you need to find where $v(t)>0$. We already did that above and found $v(t)>0$ for $t\in\left(\dfrac32,3\right)\cup(3,\infty)$.