Position Functions and Minimum Speed

Question

The position function of a particle is r(t) = <-t^2 , 5t, t^2 -4t>.

At what time is the speed minimum.

I took the derivative of the position function, then took the magnitude of that and set it equal to 0... but when I tried to do this, I couldn't isolate the t value.

If you could explain how you get to your answer, that would be helpful!

Answer 1

Given: r (t) = < -t², 5t, t² - 4t >

Find: t = ? s when || v_min || occurs

Solution:

d/dt[ r (t) ] = v (t) = < -2t, 5, 2t - 4 >

v(t)² = 4t² + 25 + 4t² - 16t + 16

v(t)² = 8t² - 16t + 41

d/dt[v(t)²] = 2 v(t) dv/dt = 0 = 16t - 16

∴ t = 1 s

v(1)² = 33 m² / s²

v(1) ≈ 5.7446 m/s

v (1) = < -2, 5, -2 > m/s

I hope this helps! Message me in the comments if you have questions, comments, or concerns!

Answer 2

The velocity vector = r'= <-2t,5,2t-4> to find the minimum take derivative of r'=r''=<-2,0,2> to find the acceleration is a constant and there will be no local minimum