Question

A particle moves along the x-axis so that at time t ≥ 0 its position is given by x(t) = 2t3 - 21t2 + 72t - 53. At what time t is the particle at rest?

Answer 1

Step-by-step explanation:

The particle is at rest when the derivative of x(t), which is its velocity, is zero, i.e.,

$\dfrac{d}{dt}[x(t)] = \dot{x}(t) = 0$

Since $x(t) = 2t^3 - 21t^2 + 72t - 53,$ its derivative is

$\dot{x}(t) = 6t^2 - 42t + 72$

Equating this to zero, we get

$6t^2 - 42t + 72 = 0 \Rightarrow t^2 - 7t + 12 = 0$

This is a familiar quadratic equation whose roots are

$t = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\;\;\;= \dfrac{7 \pm \sqrt{(7)^2 - 4(1)(12)}}{2}$

$\;\;\;= \dfrac{7 \pm 1}{2}$

$\;\;\;= 3\;\text{and}\;4$

This means that the particle will be at rest at t = 3 and t= 4.