The equation of motion of a particle is given | vedclass

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(A) The velocity $v$ is given by the derivative of displacement $x$ with respect to time $t$: $v = \frac{dx}{dt} = 50a \cos(50t + \frac{\pi}{3})$.For

Vedclass · Accepted Answer

$\frac{\pi}{300} 	ext{ s}, \frac{\pi}{75} 	ext{ s}$

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(A) The velocity $v$ is given by the derivative of displacement $x$ with respect to time $t$: $v = \frac{dx}{dt} = 50a \cos(50t + \frac{\pi}{3})$.For the particle to come to rest,$v = 0$. Thus,$\cos(50t + \frac{\pi}{3}) = 0$. The first positive value occurs when $50t + \frac{\pi}{3} = \frac{\pi}{2}$,which gives $50t = \frac{\pi}{6}$,so $t_1 = \frac{\pi}{300} 	ext{ s}$.The acceleration $a_{acc}$ is the derivative of velocity $v$ with respect to time $t$: $a_{acc} = \frac{dv}{dt} = -2500a \sin(50t + \frac{\pi}{3})$.For zero acceleration,$a_{acc} = 0$. Thus,$\sin(50t + \frac{\pi}{3}) = 0$. The first positive value occurs when $50t + \frac{\pi}{3} = \pi$,which gives $50t = \frac{2\pi}{3}$,so $t_2 = \frac{2\pi}{150} = \frac{\pi}{75} 	ext{ s}$.Therefore,$t_1 = \frac{\pi}{300} 	ext{ s}$ and $t_2 = \frac{\pi}{75} 	ext{ s}$.

The equation of motion of a particle is given by $x = a \sin(50t + \frac{\pi}{3}) \text{ cm}$. The particle will come to rest at time $t_1$ and it will have zero acceleration at time $t_2$. The $t_1$ and $t_2$ respectively are . . . . . . .

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