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The displacement of the particle from a reference point is given by √x=2t-3 .Find the distance traveled after 2 sec
According to the question;-x=(2t-3)^2Differentiating the equation w.r.t. time:-v=dx/dt=d/dt(2t-3)^2= 2(2t-3) [since,x^a=ax^a-1]we again differentiate this equation w.r.t. time;-a=dv/dt=d/dt[2(2t-3)]=4m/s^2.when t=1, v=-2m/st=4/3 , v=0m/st=2, v=2m/s V=D/T, thus D=VT;Therefore, D=2X2=4mthe distance travelled at t=2sec. is 4m.
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