$At the farthest point the velocity of particle will be zero. \ Differentiating the given equation of position and equating it with zero:$$\begin{aligned}x(t) &=5 t^{3}-9 t^{2}+7 \\x^{\prime}(t) &=15 t^{2}-18 t \\0 &=15 t^{2}-18 t \\t(15 t-18) &=0 \\t &=0 \\t &=\frac{18}{15}=1.2 \mathrm{~s}\end{aligned}$

$$$Thus, at $t=1.2 \mathrm{~s}$ particle will be at the farthest point. \\\\\underline{Justification:}\\The particle moves towards the left from $t \in[0,1.2)$. After that particle keeps moving in the positive x-direction.$

5 0

2 years ago

PLZ HELP ME!!!Davon holds a degree in sports medicine. Which company might be interested in hiring Davon?

OlgaM077 [116]

A grocery store looking for a pharmacist-

due to his degree being about medicine. I doubt the soccer team would want him as a trainer- which is the option I believe they would put to trip up the ones who don’t read the options all the way through.

4 0

2 years ago

Read 2 more answers