A particle is moving on a straight line; at time t, its position is given by the rule s(t) = t^3-75t?
Find the acceleration of the particle in ms^2 when the velocity is 0
t = secs
s(t) = metres

Question

A particle is moving on a straight line; at time t, its position is given by the rule s(t) = t^3-75t?
Find the acceleration of the particle in ms^2 when the velocity is 0
t = secs
s(t) = metres

anonymous · Answer

v(t) = s'(t)
a(t) = v'(t) = s''(t)

 helps ?

anonymous · Answer

yes i have tried this not sure that i got the right answer !

anonymous · Answer

ok what did you get? when v = 0 ? and what is the acceleration at this time?

anonymous · Answer

s'(t) = v(t) = 3t^2 -75
find when v = 0 :
0 = 3t^2 - 75
t^2 = 25 -> t =5

s''(t) = v'(t) = a(t) = 6t
a(5) = 30

anonymous · Answer

cheers thats what i got but went a bit further and confused and didnt need to cheers !!
i had square root of 900 = 30ms^2 !!!

anonymous · Answer

:)