Question

The position function of a particle is given by \({\rm{r(t) = }}\left\langle {{{\rm{t}}^{\rm{2}}}{\rm{,5t,}}{{\rm{t}}^{\rm{2}}}{\rm{ - 16t}}} \right\rangle \).When is the speed a minimum?

Short answer

The speed of the particle will be minimum when\(t = 4\).

Step-by-step solution

Speed of the particle

If\({\rm{r(t)}}\)is a differentiable vector valued function representing a position of the particle, then the speed of the particle is the magnitude of the velocity of the vector.

Finding speed of the particle

a)

Given position vector is \(r(t) = \left\langle {{t^2},5t,{t^2} - 16t} \right\rangle \).

Velocity vector can be written as-

\(\begin{aligned}{c}v(t) = r'(t)\\ = \left\langle {2t,5,2t - 16} \right\rangle \end{aligned}\)

Therefore, velocity vector is \(v(t) = \left\langle {2t,5,2t - 16} \right\rangle \).

As speed of the particle is the magnitude of the velocity;

Speed of the particle can be written as-

\(\begin{aligned}{c}|v(t)|\\ = \sqrt {{{(2t)}^2} + {{(5)}^2} + {{(2t - 16)}^2}} \\ = \sqrt {4{t^2} + 25 + 4{t^2} - 64t + 256} \\ = \sqrt {8{t^2} - 64t + 281} \\ = \sqrt {8({t^2} - 8t) + 281} \\ = \sqrt {8({t^2} - 8t + 16) + 153} \\ = \sqrt {8{{(t - 4)}^2} + 153} \end{aligned}\)

Therefore, value of speed will be minimum when \(t = 4\).