Chapter 12.2, Problem 1PP

a. If s = (2t³) m, where t is in seconds, determine v when t = 2 s.

b. If v = (5s) m/s, where s is in meters, determine a at s = 1 m.

c. If v = (4t + 5) m/s, where t is in seconds, determine a when t = 2 s.

d. If a = 2 m/s², determine v when t = 2 s if v = 0 when t = 0.

e. If a = 2 m/s², determine v at s= 4 m if v = 3 m/s at s = 0.

f. If a = (s) m/s², where s is in meters, determine v when s = 5 m if v = 0 at s = 4 m

g. If a = 4 m/s², determine s when t = 3 s if v = 2 m/s and s = 2 m when t = 0.

h. It a = (8t²) m/s², determine v when t = 1 s if v = 0 at t = 0.

i. If s = (3t² + 2) m, determine v when t = 2 s.

j. When t = 0 the particles is at A. In four seconds it travels to B, then in another six seconds it travels to C. Determine the average velocity and the average speed. The origin of the coordinate is at O.

To determine

The velocity when time is $t=2\text{s}$.

Answer to Problem 1PP

The velocity when time is $t=2\text{s}$ is $24\text{m/s}$.

Explanation of Solution

Given:

The time is $t=2\text{s}$.

The distance equation is $s=\left(2t^3\right)\text{m}$.

Write the distance equation.

$s=\left(2t^3\right)\text{m}$

Here, average velocity is ${\upsilon }_{avg}$ , change in distance is $\Delta s$ and change in time is $\Delta t$.

Write the expression velocity.

$\upsilon =\frac{ds}{dt}$ (I).

Here, velocity is $\upsilon$ and rate of change of distance $s$ with respect to time $\left(t\right)$ is $\frac{ds}{dt}$.

Conclusion:

Substitute $\left(2t^3\right)\text{m}$ for $s$ in Equation (I).

$\begin{array}{c}\upsilon =\frac{d}{dt}\left(2t^3\right)\\ =6t^2\text{m/s}\end{array}$ (II).

Substitute $2\text{s}$ for $t$ in Equation (II).

$\begin{array}{c}\upsilon =6t^2\text{m/s}\\ \text{=6}{\left(2\right)}^2\\ =24\text{m/s}\end{array}$

Thus, the velocity when time is $t=2\text{s}$ is $24\text{m/s}$.

To determine

The acceleration when distance $s=1\text{m}$.

Explanation of Solution

The acceleration $a$ when distance $s=1\text{m}$ $25{\text{m/s}}^2$.

Given:

The distance is $s=1\text{m}$.

The velocity equation is $\upsilon =\left(5s\right)\text{m/s}$.

Write the velocity equation.

$\upsilon =\left(5s\right)\text{m/s}$

Write the expression acceleration.

$a=\upsilon \frac{d\upsilon }{ds}$ (I).

Here, velocity is $\upsilon$ , acceleration is $a$ and rate of change of velocity $\upsilon$ with respect to distance $\left(s\right)$ is $\frac{d\upsilon }{ds}$.

Conclusion:

Substitute $\left(5s\right)\text{m/s}$ for $\upsilon$ in Equation (I).

$\begin{array}{c}a=\upsilon \frac{d\upsilon }{ds}\\ =\left(5s\right)\frac{d}{ds}\left(5s\right)\\ =5s\left(5\right)\\ =25s\end{array}$ (II).

Substitute $1\text{m}$ for $s$ in Equation (II).

$\begin{array}{c}a=25s\\ =25\left(1\right)\\ =25{\text{m/s}}^2\end{array}$

Thus, the acceleration $a$ when distance $s=1\text{m}$ is $25{\text{m/s}}^2$.

To determine

The acceleration when distance $t=2\text{s}$.

Answer to Problem 1PP

The acceleration $a$ when distance $t=2\text{s}$ is $4{\text{m/s}}^2$.

Explanation of Solution

Given:

The distance is $s=1\text{m}$.

The velocity equation is $\upsilon =\left(4t+5\right)\text{m/s}$.

Write the velocity equation.

$\upsilon =\left(4t+5\right)\text{m/s}$

Write the expression acceleration.

$a=\frac{d\upsilon }{dt}$ (I).

Here, acceleration is $a$ and rate of change of velocity $\upsilon$ with respect to time $\left(t\right)$ is $\frac{d\upsilon }{dt}$.

Conclusion:

Substitute $\left(4t+5\right)\text{m/s}$ for $\upsilon$ in Equation (I).

$\begin{array}{c}a=\frac{d\upsilon }{dt}\\ =\frac{d}{dt}\left(4t+5\right)\\ =4\text{m/s}\end{array}$ (II).

Thus, the acceleration $a$ when distance $t=2\text{s}$ is $4{\text{m/s}}^2$.

To determine

The velocity when time is $t=2\text{s}$.

Answer to Problem 1PP

The velocity $\upsilon$ when time is $t=2\text{s}$ is $4\text{m/s}$.

Explanation of Solution

Given:

The time is $t=2\text{s}$.

The acceleration is $a=2{\text{m/s}}^2$.

The initial velocity is ${\upsilon }_0=0$

Write the expression for final velocity in $y$ direction.

$\upsilon ={\upsilon }_0+a_{c}t$ (III).

Here, final velocity is $\upsilon$, initial velocity is ${\upsilon }_0$, acceleration due to gravity is $a_c$ and time is $t$.

Conclusion:

Substitute $2\text{s}$ for $t$, $0$ for ${\upsilon }_0$ $\left(s_0\right)$ and $2{\text{m/s}}^2$ for $a_c$ in Equation (I).

$\begin{array}{c}\upsilon ={\upsilon }_0+a_{c}t\\ =0+\left(2\right)\left(2\right)\\ =4\text{m/s}\end{array}$

Thus, the velocity $\upsilon$ when time is $t=2\text{s}$ is $4\text{m/s}$.

To determine

The velocity when distance is $s=4\text{m}$.

Answer to Problem 1PP

The velocity $\upsilon$ when distance is $s=4\text{m}$ is $5\text{m/s}$.

Explanation of Solution

Given:

The time is $t=3\text{s}$.

The acceleration is $a=2{\text{m/s}}^2$.

The initial velocity is ${\upsilon }_0=3\text{m/s}$

The initial distance is $\left(s_0\right)=0$.

The final distance is $s=4\text{m}$.

Write the expression for final velocity in $y$ direction..

${\upsilon }^2=\left({\upsilon }_0^2\right)+2a_c\left(s_y-\left(s_0\right)\right)$ (II).

Here, final velocity is $\upsilon$, initial velocity is ${\left({\upsilon }_0\right)}_y$, acceleration due to gravity is $a_c$ , final distance is $s$, initial distance is ${\left(s_0\right)}_y$.

Conclusion:

Substitute $3\text{m/s}$ for ${\upsilon }_0$ , $4\text{m}$ for $s$, $0$ for $\left(s_0\right)$ and $2{\text{m/s}}^2$ for $a_y$ in Equation (I).

$\begin{array}{l}{\upsilon }^2=\left({\upsilon }_0^2\right)+2a_c\left(s-\left(s_0\right)\right)\\ {\upsilon }^2={\left(3\right)}^2+2\left(2\right)\left(4-0\right)\\ \upsilon =5\text{m/s}\end{array}$

Thus, the velocity $\upsilon$ when distance is $s=4\text{m}$ is $5\text{m/s}$.

To determine

The velocity

Answer to Problem 1PP

The velocity $\upsilon$ is $3\text{m/s}$.

Explanation of Solution

Given:

The distance is $s_1=5\text{m}$.

The distance is $s_2=4\text{m}$.

The acceleration equation is $a=\left(s\right){\text{m/s}}^2$.

Write the acceleration equation.

$a=\left(s\right){\text{m/s}}^2$

Write the expression acceleration.

$\begin{array}{l}a=\upsilon \frac{d\upsilon }{ds}\\ ads=\upsilon d\upsilon \end{array}$ (I).

Here, velocity is $\upsilon$ , acceleration is $a$ and rate of change of velocity $\upsilon$ with respect to distance $\left(s\right)$ is $\frac{d\upsilon }{ds}$.

Conclusion:

Substitute $\left(s\right){\text{m/s}}^2$ for $a$ in Equation (I).

Integrate the Equation (I) at the limits $0$ to $\upsilon$ for $\upsilon$ and $s_1=5\text{m}$ to $s_2=4\text{m}$ for $s$.

$\begin{array}{l}{\displaystyle {\int }_0^{\upsilon }d\upsilon }={\displaystyle {\int }_4^5\left(s\right)}ds\\ {\left[\frac{{\upsilon }^2}{2}\right]}_0^{\upsilon }={\left[\frac{s^2}{2}\right]}_4^5\end{array}$

$\begin{array}{l}\frac{{\upsilon }^2}{2}=\frac{{\left(5\right)}^2}{2}-\frac{{\left(4\right)}^2}{2}\\ {\upsilon }^2=25-16\\ \upsilon =3\text{m/s}\end{array}$

Thus, the velocity $\upsilon$ is $3\text{m/s}$.

To determine

The distance when time is $t=3\text{s}$.

Answer to Problem 1PP

The distance $s$ when time is $t=3\text{s}$ is $26\text{m}$.

Explanation of Solution

Given:

The time is $t=3\text{s}$.

The acceleration is $a=4{\text{m/s}}^2$.

The velocity is $\upsilon =2\text{m/s}$

The distance is $\left(s_0\right)=2\text{m}$

Write the expression for final distance in $y$ direction.

$s=\left(s_0\right)+\left(\upsilon \right)t+\frac{1}{2}a{t}^2$ (I).

Here, final distance is $s$, initial distance is $\left(s_0\right)$, initial velocity is $\left(\upsilon \right)$, acceleration due to gravity is $a$ and time is $t$.

Conclusion:

Substitute $3\text{s}$ for $t$, $2\text{m/s}$ for $\upsilon$ , $2\text{m}$ for $\left(s_0\right)$ and $4{\text{m/s}}^2$ for $a_y$ in Equation (I).

$\begin{array}{c}s=\left(s_0\right)+\left(\upsilon \right)t+\frac{1}{2}a{t}^2\\ =2+2\left(3\right)+\frac{1}{2}\left(4\right){\left(3\right)}^2\\ =26\text{m}\end{array}$

Thus, the distance $s$ when time is $t=3\text{s}$ is $26\text{m}$.

To determine

The velocity when time is $t=1\text{s}$.

Answer to Problem 1PP

The velocity $\upsilon$ when time is $t=1\text{s}$ is $2.67\text{m/s}$.

Explanation of Solution

Given:

The time is $t=1\text{s}$.

The acceleration equation is $a=\left(8t^2\right){\text{m/s}}^2$.

Write the acceleration equation.

$a=\left(8t^2\right){\text{m/s}}^2$

Write the expression acceleration.

$\begin{array}{l}a=\frac{d\upsilon }{dt}\\ adt=d\upsilon \end{array}$ (I).

Here, acceleration is $a$ and rate of change of velocity $\upsilon$ with respect to time $\left(t\right)$ is $\frac{d\upsilon }{dt}$.

Conclusion:

Substitute $\left(8t^2\right){\text{m/s}}^2$ for $a$ in Equation (I).

Integrate the Equation (I) at the limits $0$ to $\upsilon$ for $\upsilon$ and $0$ to $t$ for $t$.

$\begin{array}{l}{\displaystyle {\int }_0^{\upsilon }d\upsilon }={\displaystyle {\int }_0^t\left(8t^2\right)}dt\\ {\left[\upsilon \right]}_0^{\upsilon }={\left[8\frac{t^3}{3}\right]}_0^t\\ \upsilon =\left(2.667t^3\right)\text{m/s}\end{array}$ (II).

Substitute $2\text{s}$ for $t$ in Equation (II).

$\begin{array}{c}\upsilon =\left(2.667t^3\right)\text{m/s}\\ =2.667\left(1\right)\\ =2.67\text{m/s}\end{array}$

Thus, the velocity when time is $t=1\text{s}$ is $2.67\text{m/s}$.

To determine

The velocity when time is $t=2\text{s}$.

Answer to Problem 1PP

The velocity $\upsilon$ when time is $t=2\text{s}$ is $12\text{m/s}$.

Explanation of Solution

Given:

The time is $t=2\text{s}$.

The distance equation is $s=\left(3t^2+2\right)\text{m}$.

Write the distance equation.

$s=\left(3t^2+2\right)\text{m}$

Here, average velocity is ${\upsilon }_{avg}$ , change in distance is $\Delta s$ and change in time is $\Delta t$.

Write the expression velocity.

$\upsilon =\frac{ds}{dt}$ (I).

Here, velocity is $\upsilon$ and rate of change of distance $s$ with respect to time $\left(t\right)$ is $\frac{ds}{dt}$.

Conclusion:

Substitute $\left(3t^2+2\right)\text{m}$ for $s$ in Equation (I).

$\begin{array}{c}\upsilon =\frac{d}{dt}\left(3t^2+2\right)\\ =6t\text{m/s}\end{array}$ (II).

Substitute $2\text{s}$ for $t$ in Equation (II).

$\begin{array}{c}\upsilon =6t\text{m/s}\\ \text{=6}\left(2\right)\\ =12\text{m/s}\end{array}$

Thus, the velocity $\upsilon$ when time is $t=2\text{s}$ is $12\text{m/s}$.

To determine

The average velocity and the average speed of the particle.

Answer to Problem 1PP

The average velocity of particle is $0.7\text{m/s}$.

The average speed of particle is $2.1\text{m/s}$.

Explanation of Solution

Given:

The distance traveled by the particle from $A$ to $C$ is shown in Figure (1).

The time traveled by the particle from $A$ to $B$ is $4\text{s}$.

The time traveled by the particle from $B$ to $C$ is $6\text{s}$.

Write the expression for the average velocity.

${\upsilon }_{avg}=\frac{\Delta s}{\Delta t}$ (I).

Here, average velocity is ${\upsilon }_{avg}$ , change in distance is $\Delta s$ and change in time is $\Delta t$.

Write the expression for the average speed ${\upsilon }_{sp}$.

${\upsilon }_{sp}=\frac{s_{Total}}{t_{Total}}$ (II).

Here, the total distance is $s_{Total}$ and the total time traveled by the particle is $t_{Total}$.

Refer Figure (1) and calculate the total distance traveled by the particle.

$s_{Total}=s_{AB}+s_{BC}$ (III)

Refer Figure (1) and calculate the total time traveled by the particle.

$t_{Total}=t_{AB}+t_{BC}$ (IV)

Conclusion:

From the Figure (1) calculate the change in distance.

$\begin{array}{c}\Delta s=\left(s_2-s_1\right)\\ =\left(6-\left(-1\right)\right)\\ =7\text{m}\end{array}$

Calculate the change in distance

$\begin{array}{c}\Delta t=\left(t_2-t_1\right)\\ =\left(10-0\right)\\ =10\text{s}\end{array}$

Substitute $10\text{s}$ for $\Delta t$ and $7\text{m}$ for $\Delta s$ in Equation (1).

$\begin{array}{c}{\upsilon }_{avg}=\frac{\Delta s}{\Delta t}\\ =\frac{7\text{m}}{10\text{s}}\\ =0.7\text{m/s}\end{array}$

Thus, the average velocity of particle is $0.7\text{m/s}$.

The time traveled by the particle from $A$ to $B$.

$t_{AB}=6\text{s}$

The time traveled by the particle from $B$ to $C$.

$t_{BC}=4\text{s}$

Substitute $7\text{m}$ for $s_{AB}$ and $14\text{m}$ for $s_{BC}$ in Equation (III).

$\begin{array}{c}{s}_{Total}=s_{AB}+s_{BC}\\ =7\text{m}\text{+14}\text{m}\\ \text{=21m}\end{array}$

Substitute $6\text{s}$ for $t_{AB}$ and $4\text{s}$ for $t_{BC}$ in Equation (IV).

$\begin{array}{c}{t}_{total}=6\text{s}+4\text{s}\\ =10\text{s}\end{array}$

Substitute $\text{21m}$ for $s_{Total}$ and $10\text{s}$ for $t_{Total}$ in Equation (II).

$\begin{array}{c}{\upsilon }_{sp}=\frac{s_{Total}}{t_{Total}}\\ =\frac{\text{21m}}{10\text{s}}\\ =2.1\text{m/s}\end{array}$

Thus, the average speed of particle is $2.1\text{m/s}$