The position of a particle moving along the x-axis is x(t)=cos(2t)-sin(3t) for time t >= 0. When t=pi, the acceleration of the particle is

Find Position Function: To find the acceleration of the particle, we need to find the second derivative of the position function x(t) with respect to time t. The position function is given by x(t) = cos(2t) - sin(3t).Find Velocity: First, we find the velocity of the particle, which is the first derivative of the position function with respect to time. We use the derivatives of cosine and sine functions, which are -sin and cos respectively, and apply the chain rule for differentiation. v(t) = dx/dt = -sin(2t) * (d/dt)(2t) - cos(3t) * (d/dt)(3t) v(t) = -2sin(2t) - 3cos(3t)Find Acceleration: Next, we find the acceleration of the particle, which is the second derivative of the position function with respect to time. We differentiate the velocity function v(t) with respect to time. a(t) = dv/dt = -2cos(2t) * (d/dt)(2t) + 3sin(3t) * (d/dt)(3t) a(t) = -4cos(2t) + 9sin(3t)Evaluate at t=pi: Now we evaluate the acceleration function at t=pi. a(pi) = -4cos(2pi) + 9sin(3pi)Substitute and Solve: We know that cos(2pi) = 1 and sin(3pi) = 0. So we substitute these values into the acceleration function. a(pi) = -4(1) + 9(0) a(pi) = -4 + 0 a(pi) = -4

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The position of a particle moving along the x-axis is $x(t)=\cos(2t)-\sin(3t)$ for time $t \geq 0$ . When $t=\pi$ , the acceleration of the particle is

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Relate position, velocity, speed, and acceleration using derivatives

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Q. The position of a particle moving along the x-axis is

x(t)=\cos(2t)-\sin(3t)

for time

t \geq 0

. When

t=\pi

, the acceleration of the particle is

Find Position Function: To find the acceleration of the particle, we need to find the second derivative of the position function $x(t)$ with respect to time $t$ . The position function is given by $x(t) = \cos(2t) - \sin(3t)$ .
Find Velocity: First, we find the velocity of the particle, which is the first derivative of the position function with respect to time. We use the derivatives of cosine and sine functions, which are $-\sin$ and $\cos$ respectively, and apply the chain rule for differentiation. $\newline$ $v(t) = \frac{dx}{dt} = -\sin(2t) \cdot \left(\frac{d}{dt}\right)(2t) - \cos(3t) \cdot \left(\frac{d}{dt}\right)(3t)$ $\newline$ $v(t) = -2\sin(2t) - 3\cos(3t)$
Find Acceleration: Next, we find the acceleration of the particle, which is the second derivative of the position function with respect to time. We differentiate the velocity function $v(t)$ with respect to time. $a(t) = \frac{dv}{dt} = -2\cos(2t) \cdot \left(\frac{d}{dt}\right)(2t) + 3\sin(3t) \cdot \left(\frac{d}{dt}\right)(3t)$ $a(t) = -4\cos(2t) + 9\sin(3t)$
Evaluate at $t=\pi$ : Now we evaluate the acceleration function at $t=\pi$ . $a(\pi) = -4\cos(2\pi) + 9\sin(3\pi)$
Substitute and Solve: We know that $\cos(2\pi) = 1$ and $\sin(3\pi) = 0$ . So we substitute these values into the acceleration function. $a(\pi) = -4(1) + 9(0)$ $a(\pi) = -4 + 0$ $a(\pi) = -4$