math(mechanics)
posted by chanz11 .
the velocity of a particle traveling in a straight line is given by v=(6t3t^2)m/s, where t is in seconds, if s=0 when t= 0. determine the particles deceleration and position when t=3s. how far has the particle traveled during the 3s time interval and what is its average speed?

if v = 6t  3t^2
then a = 6  6t
and s = distance = 3t^2  t^3 + c , where c is a constant.
but we are told that s0 when t= 0 , so c = 0
when t = 3
a = 6  6(3) =  12
s = 3(3^2)  3^3 = 0
total distance = final position  initial position
= 0  0 = 0
avg speed = 0/3 = 0
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