the velocity of a particle traveling in a straight line is given by v=(6t-3t^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?

Answer 1

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 s-0 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

Answer 2

Thanks. Very cool.