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College Physics by Openstax Chapter 2 Problem 45


A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.(a) List the knowns in this problem. (b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a known and identify its value. Then identify the unknown, and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable.(c) How long is the dolphin in the air? Neglect any effects due to his size or orientation.


Solution:

We will treat the downward direction as negative, and the upward direction as positive.

Part A

The known values are:a=-9.80\:\text{m/s}^2; v_0=13\:\text{m/s}; and y_0=0\:\text{m}.

Part B

At the highest point of the jump, the velocity is equal to 0. For this part, we will treat the initial position at the moment it jumps out of the water, and the final position at the highest point. Therefore, v_f=0 \text{m/s}.

The unknown is the final position, y_f. We are going to use the formula

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta y \\
\text{or} \\
\left(v_f\right)^2=\left(v_0\right)^2+2a\left(y_f-y_0\right)

Solving for y_f in terms of the other variables:

y_f=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2a}+y_0

Substituting the given values:

\begin{align*}
y_f & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2a}+y_0 \\
y_f & =\frac{\left(0\:\text{m/s}\right)^2-\left(13.0\:\text{m/s}\right)^2}{2\left(-9.80\:\text{m/s}^2\right)}+0\:\text{m} \\
y_f & =8.62\:\text{m}+0\:\text{m} \\
y_f & =8.62\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

This value is reasonable since dolphins can jump several meters high out of the water. Usually, a dolphin measures about 2 meters and they can jump several times their length.

Part C

The unknown is time, \Delta t. We are going to use the formula

v_f=v_0+at

Solving for time, \Delta t in terms of the other variables:

t=\frac{v_f-v_0}{a}

Substituting the given values:

\begin{align*}
t & =\frac{v_f-v_0}{a} \\
t & =\frac{0\:\text{m/s}-13.0\:\text{m/s}}{-9.80\:\text{m/s}^2} \\
t &=1.3625\:\text{s}
\end{align*}

This value is the time it takes the dolphin to reach the highest point. Since the time it takes to reach this point is equal to the time it takes to go back to the water, the time it is in the air is:

\begin{align*}
t_{air} & =2\times t \\
t_{air}&=2\times 1.3625\:\text{s} \\
t_{air}&=2.65\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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College Physics by Openstax Chapter 2 Problem 44


A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.40 m/s and observes that it takes 1.8 s to reach the water. (a) List the knowns in this problem. (b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable.


Solution:

We will treat the upward direction as positive, and the downward direction as negative.

Part A

The known values are: a=-9.80 \text{m/s}^2; v_0=-1.40\:\text{m/s}; \Delta t=1.8\:\text{s}; and y_f=0\:\text{m}

Part B

We are looking for the initial position, y_0. We are going to use the formula

\Delta y=v_{0y}t+\frac{1}{2}at^2 
\\
\text{or}
\\
y_f-y_0=v_{0y}t+\frac{1}{2}at^2

Solving for y_0 in terms of the other variables:

y_0=y_f-v_{0y}t-\frac{1}{2}at^2

Substituting the given values:

\begin{align*}
y_0 & =y_f-v_{0y}t-\frac{1}{2}at^2 \\
y_0& =0-\left(-1.4\:\text{m/s}\right)\left(1.8\:\text{s}\right)-\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)\left(1.8\:\text{s}\right)^2 \\
y_0&= 0-\left(-1.4\:\text{m/s}\right)\left(1.8\:\text{s}\right)-\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)\left(1.8\:\text{s}\right)^2 \\ 
y_0& = 0+2.52\:\text{m}+15.876\:\text{m} \\
y_0& =18.396\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

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College Physics by Openstax Chapter 2 Problem 40


(a) A world record was set for the men’s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt “coasted” across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration.

(b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?


Solution:

Part A

There are two parts to the race and must be treated separately since acceleration is not uniform over the race. We will divide the race into \Delta x_1 (while accelerating) and \Delta x_2 (with constant speed), where \Delta x_1 + \Delta x_2 = 100 \ \text{m} .

For \Delta x_1:

During the accelerating period, we are going to use the formula \Delta x=v_0t+\frac{1}{2}at^2, since we know that \displaystyle a=\frac{\Delta v}{t}=\frac{v_{max}-v_0}{t}=\frac{v_{max}}{t}; and t=3.00 \ \text{s}.

\begin{align*}
\Delta x & =v_0t+\frac{1}{2}at^2 \\
\Delta x_1 & =0+\frac{1}{2}at^2 \\
\Delta x_1 & =\frac{1}{2}at^2 \\
\Delta x_1 & =\frac{1}{2}\left(\frac{v_{max}}{t}\right)t^2 \\
\Delta x_1 & =\frac{1}{2}\left(v_{max}\right)t \\
\Delta x _1&=\frac{1}{2}\left(v_{max}\right)\left(3.00\:\text{s}\right) \\
\Delta x _1&=1.5v_{max} 
\end{align*}

When the speed is constant, t=6.69 \ \text{s}, so

\begin{align*}
\Delta x_2 & = v_{max}t \\
\Delta x_2 & = v_{max}\left(6.69\:\text{s}\right) \\
\Delta x_2 & =6.69v_{max}
\end{align*}

Plugging-in the two equations in the equation \Delta x_1 + \Delta x_2 = 100 \ \text{m} .

\begin{align*}
\Delta x_1 + \Delta x_2  & = 100 \ \text{m} \\
1.5v_{max}  + 6.69v_{max} & =100 \ \text{m} \\
8.19\:v_{max} & =100 \\
v_{max} & =\frac{100}{8.19} \\
v_{max} & =12.2\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, his acceleration can be computed using the formula

a=\frac{v_{max}}{t}

Plugging in the given values

\begin{align*}
a & =\frac{v_{max}}{t} \\
a & = \frac{12.2\:\text{m/s}}{3.00\:\text{s}} \\
a & = 4.07\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

Similar to part (a), we can plug in the different values for time and total distance:

\begin{align*}
\Delta x_1+ \Delta x_2 &  =200 \\
1.5\:v_{max}+\left(19.30-3.00\right)v_{max} & =200 \\
1.5\:v_{max}+16.30v_{max} & =200 \\
17.80v_{max} & =200 \\
v_{max} & =\frac{200}{17.80} \\
v_{max} & = 11.2\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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College Physics by Openstax Chapter 2 Problem 39


In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, of 183.58 mi/h. The one-way course was 5.00 mi long. Acceleration rates are often described by the time it takes to reach 60.0 mi/h from rest. If this time was 4.00 s, and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course?


Solution:

There are two parts to the race: an acceleration part and a constant speed part.

For the acceleration part:

We are given the following values: v_0=0 \ \text{mph} ; v_f=60 \ \text{mph}; and \Delta t=4.00 \ \text{s} .

First, we need to determine how long (both in distance and time) it takes the motorcycle to finish accelerating. During acceleration, the value of the acceleration is given by

a=\frac{60\:\text{mph}}{4\:\text{s}}

To compute for the time it takes to reach its maximum velocity, we are going to use the formula

v_f=v_0+at

Solving for time t in terms of the other variables

t=\frac{v_f-v_0}{a}

Substituting the given values to solve for t_1, the time it takes to accelerate from rest to maximum velocity:

\begin{align*}
t_1 & =\frac{v_f-v_0}{a} \\
t_1 & =\frac{183\:\text{mph}-0\:\text{mph}}{\left(\frac{60\:\text{mph}}{4\:\text{s}}\right)} \\
t_1 & =12.2\:\text{s}
\end{align*}

Since we have a constant acceleration, the distance traveled \Delta x_1 during this period is computed using the formula

\begin{align*}
\Delta x_1 & =v_{ave}t \\
\Delta x_1 & =\left(\frac{v_f+v_0}{2}\right)t \\
\end{align*}

Substituting the given values:

\begin{align*}
\Delta x_1 & =\left(\frac{v_f+v_0}{2}\right)t \\
\Delta x_1 & =\left(\frac{183\:\text{mph}+0\:\text{mph}}{2}\right)\left(12.2\:\text{s}\right) \\
\Delta x_1 & =\left(91.5\:\text{mph}\right)\left(\frac{1\:\text{hr}}{3600\:\text{s}}\right)\left(12.2\:\text{s}\right) \\
\Delta x_1 & =0.31\:\text{mi}
\end{align*}

For the constant speed part:

For the next part of the motion, the speed is constant.

We are given the following values: \Delta x_2=5.0\:\text{mi}-0.3\:\text{mi}=4.7\:\text{mi} .

We are going to solve t_2, the time spent on the course at max speed using the formula

\Delta x_2=v_{max}t_2

Solving for t_2 in terms of the other variables:

t_2=\frac{\Delta x_2}{v_{max}}

Substituting the given values:

\begin{align*}
t_2 & =\frac{\Delta x_2}{v_{max}} \\
t_2  & =\frac{4.7\:\text{mi}}{183\:\text{mph}} \\
t_2  & =\left(0.026\:\text{h}\right)\left(\frac{3600\:\text{s}}{1\:\text{h}}\right) \\
t_2  &=92\:\text{s}
\end{align*}

For the whole course:

So, the total time is

\begin{align*}
t_{total}&=t_1+t_2 \\
t_{total}& =12.2\:\text{s}+\:92\:\text{s} \\
t_{total}& =104\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}

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Problem 1-24: Calculating uncertainties in distance, time and speed of a marathon runner

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PROBLEM:

A marathon runner completes a 42.188-km course in 2 h, 30 min, and 12 s. There is an uncertainty of 25 m in the distance traveled and an uncertainty of 1 s in the elapsed time.

(a) Calculate the percent uncertainty in the distance.

(b) Calculate the uncertainty in the elapsed time.

(c) What is the average speed in meters per second?

(d) What is the uncertainty in the average speed?


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SOLUTION:

Part A

The percent uncertainty in the distance is

\begin{align*}
\text{\%\:uncertainty}_{\text{distance}} & =\frac{25\:\text{m}}{42.188\:\text{km}}\times \frac{1\:\text{km}}{1000\:\text{m}}\times 100\% \\
 & =0.0593\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Part B

The uncertainty in time is

\begin{align*}
\text{\%\:uncertainty}_{\text{time}} & =\frac{1\:\text{s}}{9012\:\text{s}}\times 100\% \\
& =0.0111\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Part C

The average speed is

\begin{align*}
\text{average speed} & =\frac{42.188\:\text{km}}{9012\:\text{s}}\times \frac{1000\:\text{m}}{1\:\text{km}} \\
& = 4.681\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Part D

The percent uncertainty in the speed is the combination of uncertainties of distance and time.

\begin{align*}
\text{\%\:uncertainty}_{\text{speed}} & =\text{\%\:uncertainty}_{\text{distance}}+\text{\%\:uncertainty}_{\text{time}} \\
& =0.0593\%+0.0111\% \\
&  =0.0704\% \\
\end{align*}

Therefore, the uncertainty in the speed is

\begin{align*}
\delta _{speed} & =\frac{0.0704\%}{100\%}\times 4.681\:\text{m/s} \\
& = 0.003\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

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