Answer
$t=4$
Work Step by Step
Need to determine the speed of a particle.
As we know $v(t)=r'(t)$ and speed is the magnitude of the velocity vector, that is ,$s(t)=|v(t)|$.
$v(t)=r'(t)=\lt 2t,5,2t-16 \gt$ [Given: $r(t)=\lt t^2, 5t, t^2-16t \gt$]
Solve for speed.
$s(t)=|v(t)|=\sqrt {(2t)^2+(5)^2+( 2t-16)^2}$
$s(t)=\sqrt {4t^2+25+4t^2+256-64t}$
$s(t)=\sqrt {8t^2-64t+281}$
Our aim is to find the minimum value of speed, in order to this take the extreme value of the quadratic equation.
Simplify to get thje value of $t$.
$t=-\dfrac{b}{2a}=-\dfrac{(-64)}{2(8)}=4$