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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
Solution
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.