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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]

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  • 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$


answered Jul 22, 2013 by meena.p

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