Chapter 3.3, Problem 35E

(a)

To determine

To find: The velocity and acceleration at time t.

Answer to Problem 35E

The velocity of smooth level surface is $\underset{\_}{8\cos t}$.

The acceleration of smooth level surface is $\underset{\_}{-8\sin t}$.

Explanation of Solution

Given:

The equation of motion is $x\left(t\right)=8\sin t$, where t is in seconds and x is in centimeters.

Derivative rule:

Constant Multiple Rule:

If c is constant and $f\left(x\right)$. is differentiable function, then

$\frac{d}{dx}\left[cf\left(x\right)\right]=c\frac{d}{dx}\left[f\left(x\right)\right]$ (1)

Recall:

If $x\left(t\right)$x is a displacement of a particle and the time t is in seconds, then the velocity of the particle is $v\left(t\right)=\frac{dx}{dt}$.

If $v\left(t\right)$ is a velocity of the particle and the time t is in seconds, then the acceleration of the particle is $a\left(t\right)=\frac{dv}{dt}$.

Calculation:

Obtain the velocity at time t.

$\begin{array}{c}v\left(t\right)=\frac{d}{dt}\left(x\left(t\right)\right)\\ =\frac{d}{dt}\left(8\sin t\right)\end{array}$

Apply the Constant Multiple Rule as shown in equation (1),

$\begin{array}{c}v\left(t\right)=8\frac{d}{dx}\left(\sin t\right)\\ =8\cos t\end{array}$

Thus, the velocity of $x\left(t\right)=8\sin t$ is $v\left(t\right)=8\cos t$.

Obtain the acceleration at time t.

$\begin{array}{c}a\left(t\right)=\frac{d}{dt}\left(v\left(t\right)\right)\\ =\frac{d}{dt}\left(8\cos t\right)\end{array}$

Apply the Constant Multiple Rule as shown in equation (1),

$\begin{array}{c}a\left(t\right)=8\frac{d}{dx}\left(\cos t\right)\\ =8\left(-\sin t\right)\\ =-8\sin t\end{array}$

Therefore, the acceleration of $v\left(t\right)=8\cos t$ is $a\left(t\right)=-8\sin t$.

(b)

To determine

To find: The position, velocity and acceleration at $t=\frac{2\pi }{3}$ and obtain its direction.

Answer to Problem 35E

The position, velocity and acceleration at $t=\frac{2\pi }{3}$ are $4\sqrt{3},-\text{4}$ and $-4\sqrt{3}$, respectively.

The direction of the particle is moving to the left (negative direction).

Explanation of Solution

Given:

The equation of motion is $x\left(t\right)=8\sin t$, where t is in seconds and x is in centimeters.

Calculation:

The position $x\left(t\right)=8\sin t$ at $t=\frac{2\pi }{3}$ is computed as follows,

$\begin{array}{c}x\left(\frac{2\pi }{3}\right)=8\sin \frac{2\pi }{3}\\ =8\left(\frac{\sqrt{3}}{2}\right)\left(Q\sin \frac{2\pi }{3}\text{=}\frac{\sqrt{3}}{2}\right)\\ =4\sqrt{3}\end{array}$

Thus, the position $x\left(t\right)=8\sin t$ at $t=\frac{2\pi }{3}$ is $x\left(\frac{2\pi }{3}\right)=4\sqrt{3}$.

From part (a), the velocity of $x\left(t\right)$ is $v\left(t\right)=8\cos t$.

The velocity $v\left(t\right)=8\cos t$ at $t=\frac{2\pi }{3}$ is computed as follows,

$\begin{array}{c}v\left(\frac{2\pi }{3}\right)=8\cos \left(\frac{2\pi }{3}\right)\\ =8\left(-\frac{1}{2}\right)\left(Q\cos \frac{2\pi }{3}=-\frac{1}{2}\right)\\ =-4\end{array}$

Thus, the velocity $v\left(t\right)=8\cos t$ at $t=\frac{2\pi }{3}$ is $v\left(\frac{2\pi }{2}\right)=-4$.

From part (a), the acceleration of $v\left(t\right)=8\cos t$ is $a\left(t\right)=-8\sin t$.

The acceleration $a\left(t\right)=-8\sin t$ at $t=\frac{2\pi }{3}$ is computed as follows,

$\begin{array}{c}a\left(\frac{2\pi }{3}\right)=-8\sin \frac{2\pi }{3}\\ =-8\left(\frac{\sqrt{3}}{2}\right) \left(Q\sin \frac{2\pi }{3}=\frac{\sqrt{3}}{2}\right)\\ =-4\sqrt{3}\end{array}$

Thus, the acceleration $a\left(t\right)=-8\sin t$ at $t=\frac{2\pi }{3}$ is $a\left(\frac{2\pi }{3}\right)=-4\sqrt{3}$.

Here, the velocity of the particle at $t=\frac{2\pi }{3}$ is less than zero. That is, $v\left(\frac{2\pi }{3}\right)<0$.

This implies that, the velocity of the particle at $t=\frac{2\pi }{3}$ is negative.

Therefore,the direction of the particle is moving to the left (negative direction).