The position function of a particle is given below. s(t) = t^3 + 3t^2 − 69t, t ≥ 0 When does the particle reach a velocity of 3 m/s?

Answer 1

s = t^3 + 3t^2 -69t

v= s' = 3t^2 +6t -69 = 3

t^2 +2t -23 =1

t^2 +2t -24 =0

(t+6)(t-4) = 0

t-4=0

t = 4 seconds when v=3 m/sec

Answer 2

s(t) = t^3 + 3t^2 - 69t

Velocity is the derivative of the position, so v(t) = s'(t) = 3t^2 + 6t - 69 = 3(t^2 + 2t - 23)

To find when velocity is 3 m/s, set the velocity equation equal to 3 and solve for t:

3(t^2 + 2t - 23) = 3 Divide both sides by 3

t^2 + 2t - 23 = 1 Subtract 1 from both sides

t^2 + 2t - 24 = 0 Now factor

(t + 6)(t - 4) = 0 Set each factor equal to zero

t + 6 = 0 t - 4 = 0

t = -6 t = 4 We ignore the solution of t = -6 since t must be 0 or greater.

The solution: The particle reaches a velocity of 3 m/s at t = 4 seconds into its journey.