physics
posted by j on .
the motion of a particle along a straight line is described by the function x=(2t3)^2 where x is in metres and t is in seconds.
A)find the position ,veocity and acceleration at t=2 sec.
B) find the velocity of the particle at origin.

A) position: x(2)= (2*23)^2= 1m
v = dx/dt = 2(2t3)*2
so v(2)= 4 m/s
a = dv/dt = 8 m/s^2
B) v at origin: x=0=(2t3)^2
so t=3/2
at t=3/2, v = 4(2*3/23)= 0