Chapter 2.1, Problem 8E

The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s = 2 sin πt + 3 cos πt, where t is measured in seconds.

(a) Find the average velocity during each time period:

(i) [1, 2]

(ii) (1, 1.1]

(iii) [1, 1.01]

(iv) [1, 1.001]

(b) Estimate the instantaneous velocity of the particle when t = 1.

To determine

To find: The average velocity for given time periods.

Answer to Problem 8E

(i) The average velocity over the time interval [1, 2] is $6\text{ cm/s}$.

(ii) The average velocity over the time interval [1, 1.1] is $-4.71\text{ cm/s}$.

(iii) The average velocity over the time interval [1, 1.01] is $-6.13\text{ cm/s}$.

(iv) The average velocity over the time interval [1, 1.001] is $-6.27\text{ cm/s}$.

Explanation of Solution

Given:

The equation of a motion $s=2\sin \pi t+3\cos \pi t$ (in centimeters), where t is measured in seconds.

Formula used:

The average velocity over the time interval $\left[t,t+h\right]$ is,

$v_{avg}=\frac{s\left(t+h\right)-s\left(t\right)}{\left(t+h\right)-t}$ (1)

Calculation:

Section-(i)

Obtain the average velocity over the time interval [1, 2].

Substitute $t=1$ and $t+h=2$ in equation (1),

$\begin{array}{c}{v}_{\text{avg}}=\frac{s\left(t+h\right)-s\left(t\right)}{\left(t+h\right)-t}\\ =\frac{s\left(2\right)-s\left(1\right)}{2-1}\\ =\frac{\left(2\sin \pi \left(2\right)+3\cos \pi \left(2\right)\right)-\left(2\sin \pi \left(1\right)+3\cos \pi \left(1\right)\right)}{1}\\ =2\sin 2\pi +3\cos 2\pi -2\sin \pi -3\cos \pi \end{array}$

Substitute the respective trigonometric values and simplify the terms.

$\begin{array}{c}{v}_{\text{avg}}=2\left(0\right)+3\left(1\right)-2\left(0\right)-3\left(-1\right)\\ =0+3-0+3\\ =6\end{array}$

Thus, the average velocity over the time interval [1, 2] is $6\text{ cm/s}$.

Section-(ii)

Obtain the average velocity over the time interval [1, 1.1].

Substitute $t=1$ and $t+h=1.1$ in equation (1),

$\begin{array}{c}{v}_{\text{avg}}=\frac{s\left(t+h\right)-s\left(t\right)}{\left(t+h\right)-t}\\ =\frac{s\left(1.1\right)-s\left(1\right)}{1.1-1}\\ =\frac{\left(2\sin \pi \left(1.1\right)+3\cos \pi \left(1.1\right)\right)-\left(2\sin \pi \left(1\right)+3\cos \pi \left(1\right)\right)}{0.1}\\ =\frac{2\sin \left(1.1\pi \right)+3\cos \left(1.1\pi \right)-2\sin \pi -3\cos \pi }{0.1}\end{array}$

Substitute the respective trigonometric values and simplify the terms.

$\begin{array}{c}{v}_{\text{avg}}=\frac{2\left(-0.30902\right)+3\left(-0.9511\right)-2\left(0\right)-3\left(-1\right)}{0.1}\\ =\frac{-0.61804-2.8533+3}{0.1}\\ =\frac{0.47134}{0.1}\\ \approx -4.71\end{array}$

Thus, the average velocity over the time interval [1, 1.1] is $-4.71\text{ cm/s}$.

Section-(iii)

Obtain the average velocity over the time interval [1, 1.01].

Substitute $t=1$ and $t+h=1.01$ in equation (1),

$\begin{array}{c}{v}_{\text{avg}}=\frac{s\left(t+h\right)-s\left(t\right)}{\left(t+h\right)-t}\\ =\frac{s\left(1.01\right)-s\left(1\right)}{1.01-1}\\ =\frac{\left(2\sin \pi \left(1.01\right)+3\cos \pi \left(1.01\right)\right)-\left(2\sin \pi \left(1\right)+3\cos \pi \left(1\right)\right)}{0.01}\\ =\frac{2\sin \left(1.01\pi \right)+3\cos \left(1.01\pi \right)-2\sin \pi -3\cos \pi }{0.01}\end{array}$

Substitute the respective trigonometric values and simplify the terms.

$\begin{array}{c}{v}_{\text{avg}}=\frac{2\left(-0.03141\right)+3\left(-0.9995\right)-2\left(0\right)-3\left(-1\right)}{0.01}\\ =\frac{-0.06282-2.9985+3}{0.01}\\ =\frac{-0.06132}{0.01}\\ \approx -6.13\end{array}$

Thus, the average velocity over the time interval [1, 1.01] is $-6.13\text{ cm/s}$.

Section-(iv)

Obtain the average velocity over the time interval [1, 1.001].

Substitute $t=1$ and $t+h=1.001$ in equation (1).

$\begin{array}{c}{v}_{\text{avg}}=\frac{s\left(t+h\right)-s\left(t\right)}{\left(t+h\right)-t}\\ =\frac{s\left(1.001\right)-s\left(1\right)}{1.001-1}\\ =\frac{\left(2\sin \pi \left(1.001\right)+3\cos \pi \left(1.001\right)\right)-\left(2\sin \pi \left(1\right)+3\cos \pi \left(1\right)\right)}{0.001}\\ =\frac{2\sin \left(1.001\pi \right)+3\cos \left(1.001\pi \right)-2\sin \pi -3\cos \pi }{0.001}\end{array}$

Substitute the respective trigonometric values and simplify the terms.

$\begin{array}{c}{v}_{\text{avg}}=\frac{2\left(-0.0031416\right)+3\left(-0.999995\right)-2\left(0\right)-3\left(-1\right)}{0.001}\\ =\frac{-0.0062832-2.999985+3}{0.001}\\ =\frac{-0.0062682}{0.001}\\ \approx -6.27\end{array}$

Thus, the average velocity over the time interval [1, 1.001] is $-6.27\text{ cm/s}$.

To determine

To estimate: The instantaneous velocity of the particle when $t=1$.

Answer to Problem 8E

The estimated instantaneous velocity when $t=1$ is $-6.3\text{ cm/s}$.

Explanation of Solution

The average velocity over the time interval [1, 1+h] is,

$\begin{array}{c}{v}_{avg}=\frac{s\left(1+h\right)-s\left(1\right)}{\left(1+h\right)-1}\\ =\frac{2\sin \pi \left(1+h\right)+3\cos \pi \left(1+h\right)-\left(2\sin \pi \left(1\right)+3\cos \pi \left(1\right)\right)}{h}\\ =\frac{2\sin \pi \left(1+h\right)+3\cos \pi \left(1+h\right)-\left(2\left(0\right)+3\left(-1\right)\right)}{h}\\ =\frac{2\sin \pi \left(1+h\right)+3\cos \pi \left(1+h\right)+3}{h}\end{array}$

The instantaneous velocity when $t=1$ is expressed as the limiting value of the average velocities over smaller time periods that start at $t=1$.

That is, when h approaches 0, then t is closer to 1.

$\begin{array}{c}v=\underset{h\to 0}{\lim }\frac{s\left(1+h\right)-s\left(1\right)}{h}\\ =\underset{h\to 0}{\lim }\left(\frac{2\sin \pi \left(1+h\right)+3\cos \pi \left(1+h\right)+3}{h}\right)\\ =-6.2831\\ \approx -6.3\end{array}$

Thus, the estimated instantaneous velocity when $t=1$ is $-6.3\text{ cm/s}$.