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

Explanation:

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

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