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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.

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

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2 answers

  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

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