Answer
$t=4$
Work Step by Step
Given: $r(t)=\lt t^2, 5t, t^2-16t \gt$
In order to calculate the speed of a particle, we will take the help of the following formulas such as:
$v(t)=r'(t)$ and speed is the magnitude of the velocity vector, that is, $s(t)=|v(t)|$.
Now,
$v(t)=r'(t)=\lt 2t,5,2t-16 \gt$
Speed: $s(t)=|v(t)|=\sqrt {(2t)^2+(5)^2+( 2t-16)^2}=\sqrt {4t^2+25+4t^2+256-64t}$
$s(t)=\sqrt {8t^2-64t+281}$
To calculate the minimum value of speed we will have to take the extreme value of the quadratic equation. The expression for speed shows a quadratic equation under the square root, whose minimum value can be calculated as: $t=-\dfrac{b}{2a}=-\dfrac{(-64)}{2(8)}=4$
Hence, the required answer is $t=4$
