Calculus question: (Do the anaylsis with calc) A particle moves on the x axis according to the function s(t)=t^2-8t+15 at time t seconds.

Question

A particle moves on the x axis according to the function s(t)=t^2-8t+15 at time t seconds.

a) At what time is it moving forward? backward?

b) when is the speed increasing

c) when is the particle stopped

Answer 1

velocity v(t) = s'(t) = 2t-8, critical point: t= 4

a(t) = v '(t) = 2

s(4) = -1 which can be shown to be a minimum, s(3) = 0 and s(5) = 0 both greater than -1

a) Velocity is greater than zero when t > 4. It is moving backward when t < 4.

b) Speed is increasing when t> 4

c) The particle is stopped at the critical point when t = 4.