Question

Particle x moves along the positive x-axis so that its position at time t ~ 0 is given by x(t) = 5t 3 - 9t 2 + 7. at what time t ~ 0 is particle x farthest to the left? justify your answer

Answer 1

Answer: 1.2 seconds

Step-by-step explanation:

`$$At the farthest point the velocity of particle will be zero. \ Differentiating the given equation of position and equating it with zero:$$\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 &=(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.`