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# A particle of unit mass moves so that displacement after $t$ secs is given by $x=3\cos(2t-4).$ Find the acceleration and kinetic energy at the end of $2 \;secs.[ K.E =\large\frac{1}{2}$$mv^{2}\quad m is mass] Can you answer this question? ## 1 Answer 0 votes Toolbox: • If s=f(t) is the distance function, representing the distance 's' travelled by a particle in time t, then the velocity and acceleration functions are v=\large\frac{ds}{dt}$$=t'(t)$ and $a=\large\frac{d^2s}{dt^2}$$=f''(t) x=3 \cos (2t-4)\; -----(i) mass=1 Step 1: Acceleration is given by \large\frac{d^2x}{dt^2} Step 2: Differentiating (i) twice, we get \large\frac{dx}{dt}$$=3(-2s \sin (2t-4))=-6 \sin (2t-4)$----(i)
$\large\frac{d^2x}{dt^2}$$=-6 \times 2 \cos (2t-4)=-12 \cos (2t-4)----(ii) Step 3: When t=2 Acceleration =-12\; \cos 0=-12 The velocity after 2 sec is got by substituting t=2 in (i) Velocity v=-6 \sin 0=0 Therefore the K.E= \large\frac{1}{2}$$mv^2=0$

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