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(B) The displacement of the body is given by $x = 3 \cos \left( 2 \pi t + \frac{\pi}{4} \right)$.Velocity $v$ is the first derivative of displacement

Vedclass · Accepted Answer

$-6 \sqrt{2} \pi^2 	ext{ m/s}^2$

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(B) The displacement of the body is given by $x = 3 \cos \left( 2 \pi t + \frac{\pi}{4} ight)$.Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{dx}{dt} = -3 \sin \left( 2 \pi t + \frac{\pi}{4} ight) \cdot (2 \pi) = -6 \pi \sin \left( 2 \pi t + \frac{\pi}{4} ight)$.Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = -6 \pi \cos \left( 2 \pi t + \frac{\pi}{4} ight) \cdot (2 \pi) = -12 \pi^2 \cos \left( 2 \pi t + \frac{\pi}{4} ight)$.At $t = 2 	ext{ s}$, the acceleration is $a = -12 \pi^2 \cos \left( 2 \pi (2) + \frac{\pi}{4} ight) = -12 \pi^2 \cos \left( 4 \pi + \frac{\pi}{4} ight)$.Since $\cos(4 \pi + 	heta) = \cos 	heta$, we have $a = -12 \pi^2 \cos \left( \frac{\pi}{4} ight) = -12 \pi^2 \cdot \frac{1}{\sqrt{2}} = -6 \sqrt{2} \pi^2 	ext{ m/s}^2$.

If the displacement of a body is given by $x = 3 \cos \left( 2 \pi t + \frac{\pi}{4} \right) \text{ m}$, then the acceleration of the body at $t = 2 \text{ s}$ is

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