# math

A particle moves along straight line such that its displacement S meters from a given point is S = t^3 – 5t^2 + 4 whee t is time in seconds.
Find
(a) The displacement of particle at t = 5
(b) The velocity of the particle when t = 5
(c) The values of t when the particle is momentarily at rest
(d) The acceleration of the particle when t = 2

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1. S(5) = 125 -125 + 4 = 4

v = 3 t^2 - 10 t
v(5) = 75 - 50

when is v = 0?
3 t^2 - 10 t = 0 = t(3 t - 10)
t = 0 or t - 10/3

a = 6 t - 10
a(2) = 12 - 10 = 2

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posted by Damon

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