Question

Finding Speed A body's velocity at time \(t\) sec is \(v=2 t^{3}-9 t^{2}+12 t-5 \mathrm{m} / \mathrm{sec} .\) Find the body's speed each time the acceleration is zero.

Short answer

The body's speed when acceleration is zero is 0 m/sec at \(t=1 sec\) and 1 m/sec at \(t=2 sec\).

Step-by-step solution

Compute the derivative of the velocity function

The derivative of the velocity function \(v(t) =2 t^{3}-9 t^{2}+12 t-5\) is \(v'(t) =6 t^{2}-18 t+12\). This derivative represents the acceleration of the body.

Find the time when the acceleration is zero

Set the derivative equal to zero, \(0=6 t^{2}-18 t+12\). This can be simplified by taking out a common factor of 6 to get \(0=t^{2}-3 t+2\). Factoring this gives \(0=(t-1)(t-2)\). Setting each factor to zero gives the time as \(t=1\) and \(t=2\) sec when the acceleration is zero.

Find the speed when acceleration is zero

Substitute each time into the velocity equation to find the speed: at \(t=1 sec\), \(v(1) =2(1)^{3}-9(1)^{2}+12(1)-5 = 0 \), and at \(t=2 sec\), \(v(2) =2(2)^{3}-9(2)^{2}+12(2)-5 = 1 \) m/sec. So the body’s speed when the acceleration is zero is 0 m/sec at \(t=1 sec\) and 1 m/sec at \(t=2 sec\).