A particle moves along the x-axis with velocity given by v(t)=12 sin(4t) for time t >= 0. If the particle is at position x=1 at time t=0, what is the position of the particle at time t=(pi)/(8) ?

Set up integral for position: To find the position of the particle at time t=(pi)/(8), we need to integrate the velocity function v(t)=12 sin(4t) from t=0 to t=(pi)/(8). The position function x(t) is the antiderivative of the velocity function v(t), plus the initial position x(0)=1.Integrate velocity function: First, we set up the integral of the velocity function to find the position function x(t): x(t) = ∫ v(t) dt from 0 to t + x(0) x(t) = ∫ 12 sin(4t) dt from 0 to t + 1Find constant of integration: Now we integrate 12 sin(4t) with respect to t: The antiderivative of sin(4t) is -(1/4)cos(4t), so the antiderivative of 12 sin(4t) is -3cos(4t). x(t) = -3cos(4t) + C, where C is the constant of integration.Calculate position function: We find the constant of integration C by using the initial condition x(0)=1: x(0) = -3cos(4*0) + C = 1 C = 1 + 3cos(0) C = 1 + 3(1) C = 4Evaluate position at t=(pi)/(8): Now we have the position function x(t): x(t) = -3cos(4t) + 4We evaluate the position function at t=(pi)/(8): x((pi)/(8)) = -3cos(4*(pi)/(8)) + 4 x((pi)/(8)) = -3cos((pi)/(2)) + 4 Since cos((pi)/(2)) = 0, the equation simplifies to: x((pi)/(8)) = -3(0) + 4 x((pi)/(8)) = 4

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A particle moves along the
x-axis with velocity given by
v(t)=12 sin(4t) for time
t >= 0. If the particle is at position
x=1 at time
t=0, what is the position of the particle at time
t=(pi)/(8) ?

A particle moves along the $x$ -axis with velocity given by $v(t)=12 \sin (4 t)$ for time $t \geq 0$ . If the particle is at position $x=1$ at time $t=0$ , what is the position of the particle at time $t=\frac{\pi}{8}$ ?

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Calculus

Relate position, velocity, speed, and acceleration using derivatives

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Q. A particle moves along the

x

-axis with velocity given by

v(t)=12 \sin (4 t)

for time

t \geq 0

. If the particle is at position

x=1

at time

t=0

, what is the position of the particle at time

t=\frac{\pi}{8}

Set up integral for position: To find the position of the particle at time $t=\frac{\pi}{8}$ , we need to integrate the velocity function $v(t)=12 \sin(4t)$ from $t=0$ to $t=\frac{\pi}{8}$ . The position function $x(t)$ is the antiderivative of the velocity function $v(t)$ , plus the initial position $x(0)=1$ .
Integrate velocity function: First, we set up the integral of the velocity function to find the position function $x(t)$ :
$x(t) = \int_{0}^{t} v(t) \, dt + x(0)$
$x(t) = \int_{0}^{t} 12 \sin(4t) \, dt + 1$
Find constant of integration: Now we integrate $12 \sin(4t)$ with respect to $t$ : The antiderivative of $\sin(4t)$ is $-(1/4)\cos(4t)$ , so the antiderivative of $12 \sin(4t)$ is $-3\cos(4t)$ . $x(t) = -3\cos(4t) + C$ , where $C$ is the constant of integration.
Calculate position function: We find the constant of integration $C$ by using the initial condition $x(0)=1$ : $x(0) = -3\cos(4\cdot 0) + C = 1$ $C = 1 + 3\cos(0)$ $C = 1 + 3(1)$ $C = 4$
Evaluate position at $t=\frac{\pi}{8}$ : Now we have the position function $x(t)$ : $\newline$ $x(t) = -3\cos(4t) + 4$
Evaluate position at $t=\frac{\pi}{8}$ : Now we have the position function $x(t)$ : $x(t) = -3\cos(4t) + 4$ We evaluate the position function at $t=\frac{\pi}{8}$ : $x\left(\frac{\pi}{8}\right) = -3\cos\left(4\cdot\frac{\pi}{8}\right) + 4$ $x\left(\frac{\pi}{8}\right) = -3\cos\left(\frac{\pi}{2}\right) + 4$ Since $\cos\left(\frac{\pi}{2}\right) = 0$ , the equation simplifies to: $x\left(\frac{\pi}{8}\right) = -3(0) + 4$ $x\left(\frac{\pi}{8}\right) = 4$