Home Page Textbooks Physics Page 59: Practice Problems

Physics

1st Edition

Walker

ISBN: 9780133256925

Textbook solutions

All Solutions

Page 59: Practice Problems

Exercise 35

Step 1

1 of 3

$textbf{(a)}$ Using the $textbf{position-time equation of motion}$

$$
begin{align*}
x_{f}=x_{i}+vt\
end{align*}
$$

where $x_{f}$ is the final position, $x_{i}$ initial position, $v$ velocity, and $t$ elapsed time, we can quickly determine that the bunny’s initial position is

$$
begin{align*}
boxed{x_{i}=8.3 text{m}}\
end{align*}
$$

Step 2

2 of 3

$textbf{(b)}$ Using the abovementioned equation of motion, we come to conclusion that the bunny’s velocity is

$$
begin{align*}
boxed{v=2.2 frac{text{m}}{text{s}}}\
end{align*}
$$

Result

3 of 3

$$
begin{align*}
textbf{(a)} quad &boxed{x_{i}=8.3 text{m}}\
\
textbf{(b)} quad &boxed{v=2.2 frac{text{m}}{text{s}}}\
end{align*}
$$

Exercise 36

Solution 1

Solution 2

Solution 3

Step 1

1 of 5

$textbf{(a)}$ The objective is to write the position-time equation for the ball, given its initial and final position, as well as the elapsed time.

Provided that the initial position and the velocity of an object are known, the position-time equation gives its position $x$ at any time $t$,

$$
begin{align*}
x_{f}=x_{i}+vt
end{align*}
$$

This means that $textbf{we have to calculate the ball’s velocity first}$ if we are to write the position-time equation.

Step 2

2 of 5

Using $textbf{the formula for average velocity}$ and plugging in the values, we have:

$$
begin{align*}
v&=dfrac{Delta x}{Delta t}=dfrac{x_{f}-x_{i}}{Delta t}\
&=dfrac{7.8 text{m}-1.6 text{m}}{3.1 text{s}}\
&=quadboxed{2.0 dfrac{text{m}}{text{s}}}\
end{align*}
$$

Substituting the values for the initial position and the velocity in the equation of motion gives:

$$
begin{align*}
boxed{x_{f}=1.6 text{m}+left(2.0 dfrac{text{m}}{text{s}}right)t}\
end{align*}
$$

Step 3

3 of 5

$textbf{(b)}$ In this part, $textbf{the goal is to find the time $t$ at which the ball is at the position given in the question.}$ Substituting $8.6 text{m}$ for the final position into the equation of motion gives the corresponding time:

$$
begin{align*}
8.6 text{m}&=1.6 text{m}+left(2.0 dfrac{text{m}}{text{s}}right)t\
rightarrowquad t&=dfrac{8.6 text{m}-1.6 text{m}}{2.0 frac{text{m}}{text{s}}}\
&=quadboxed{3.5 text{s}}\
end{align*}
$$

Step 4

4 of 5

* Notice that we applied $textbf{the rule for multiplication and division for significant figures}$ when writing the final results for $v$ and $t$.

Result

5 of 5

$$
begin{align*}
textbf{(a)}quad &boxed{x_{f}=1.6 text{m}+left(2.0 dfrac{text{m}}{text{s}}right)t}\
\
\
\
textbf{(b)}quad &boxed{t=3.5 text{s}}\
end{align*}
$$

Step 1

1 of 3

$$
tt {(a)To find the velocity we Rearrange the Position-Time equation of Motion with $v$ as a subject:}
$$

$$
begin{align*}
x_f=x_i+vt Rightarrow v &= frac{x_f-x_i}{t}\
&=frac{(7.8m)-(1.6m)}{(3.1s)}\
&= boxed{2 frac{m}{s})}
end{align*}
$$

$tt {Finally we substitute the variables of the equation with the actual values:}$

$$
boxed{textcolor{#4257b2}{x_f=(1.6m)+(2frac{m}{s})t}}
$$

Step 2

2 of 3

$$
tt{ (b) Rearrange the Position-Time equation of Motion with $t$ as a subject:}
$$

$$
begin{align*}
t &= frac{x_f-x_i}{v}\
&=frac{(8.6m)-(1.6m)}{(2frac{m}{s})}\
&=boxed{textcolor{#4257b2}{3.5m}}
end{align*}
$$

Result

3 of 3

$tt {(a) $x_f =(1.6m)+(2frac{m}{s})t$}$. (b) $3.5m$

Step 1

1 of 3

a.) $textbf{Concept:}$
The equation in the linear form look like $x_f=x_i+vt$ where time $t$ is multiplied by velocity $v$, and the answer is added to the initial position. We will use the given equation to find the velocity together with initial speed of the bowling ball.

$$
textbf{Solution:}
$$

The displacement divided by times gives the velocity of the bowl:

$$
v_{av}=frac{Delta x}{Delta t}=frac{7.8m-1.6m}{3.1s}=color{#4257b2} boxed{bf 2.0m/s}
$$

Now writing the equation in the form $x_f=x_i+vt$ we get

$$
x_f=color{#4257b2} boxed{bf 1.6m+2.0m/stimes t}
$$

Step 2

2 of 3

b.) $textbf{Concept:}$
Computing $t$ from the equation we get

$$
textbf{Solution:}
$$

$$
x_f-x_i=vt
$$

$$
frac{x_f-x_i}{v}=t
$$

Substituting values we get

$$
t=frac{8.6-1.6m}{20m/s}=color{#4257b2} boxed{bf 3.5s}
$$

Result

3 of 3

$$
x_f=1.6m+2.0m/stimes t;and;;x_f=1.6m+2.0m/stimes t
$$

unlock

All Solutions

Page 59: Practice Problems

Haven't found what you were looking for?

Search for samples, answers to your questions and flashcards