Question

A particle moves along the x-axis so that at time t\ge 0t≥0 its velocity is given by v(t)=2t-7.v(t)=2t−7. Determine if the particle is speeding up or slowing down at t=9.t=9.

Answer 1

SOLUTION

$\begin{gathered} v(t)\text{ = 2t - 7} \\ v(9)\text{ = 2(9) - 7} \\ v(9)\text{ = 18 - 7} \\ v(9)\text{ = 11} \end{gathered}$

To find a(9), we differentiate or find the derivative of v(t)

$\begin{gathered} v(t)\text{ = 2t - 7} \\ \frac{d\text{ v}}{d\text{ t}}\text{ = 2} \\ \\ a(9)\text{ = 2} \end{gathered}$

The velocity is increasing, while the acceleration is constant