particles and velocity question

Question

The position function of a particle is given below.

s(t) = t^3 + 3t^2 -19t,t>=0

When does the particle reach a velocity of 5 m/s?

t= ?

Answer 1

v(t) = ds/dt = d/dt(t³ +3t²-19t) = 3t²+6t-19

So, solve for t that yields v(t) = 5 m/s: 5 = 3t²+6t-19 or 3t² + 6t - 24

t = (-6 ± sqrt( 36 +4*24*3))/(2*3) = -1 ± 18/6 = -4, 2 seconds.

Normally you would throw out the negative solution for time, but it is not entirely clear here as we may have started a timer at t =0, but we might be interested in when the particle went that fast before we started timing (given that the position function applied in the past)

Answer 2

velocity v(t) = s'(t) = 3t² + 6t - 19. To get time t we set v(t) = 5m/s;

WE have equation: 3t² + 6t - 19 = 5; 3t² + 6t -24 = 0, t² + 2t - 8 = 0; (t + 4)(t - 2) = 0, t = -4 or t = 2.

Answer: t = 2sec