the motion of a particle along a straight line is described by the function x=(2t-3)^2 where x is in metres and t is in seconds.

Question

A)find the position ,veocity and acceleration at t=2 sec.
B) find the velocity of the particle at origin.

Answer 1

A) position: x(2)= (2*2-3)^2= 1m
v = dx/dt = 2(2t-3)*2
so v(2)= 4 m/s

a = dv/dt = 8 m/s^2

B) v at origin: x=0=(2t-3)^2
so t=3/2
at t=3/2, v = 4(2*3/2-3)= 0

Answer 2

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