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College Physics by Openstax Chapter 2 Problem 32

A woodpecker’s brain is specially protected from large decelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker’s head comes to a stop from an initial velocity of 0.600 m/s in a distance of only 2.00 mm.

a) Find the acceleration in m/s2 and in multiples of g (g=9.80 m/s2).

b) Calculate the stopping time.

c) The tendons cradling the brain stretch, making its stopping distance 4.50 mm (greater than the head and, hence, less deceleration of the brain). What is the brain’s deceleration, expressed in multiples of g?

Solution:

We are given the following values: v_0=0.600 \ \text{m/s}; v_f=0.000\:\text{m/s}; and \Delta x=0.002\:\text{m}.

Part A

The acceleration is computed based on the formula, 

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Solving for acceleration a in terms of the other variables, we have

a=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substituting the given values, 

\begin{align*}
a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\
a & =\frac{\left(0.000\:\text{m/s}\right)^2-\left(0.600\:\text{m/s}\right)^2}{2\left(0.002\:\text{m}\right)} \\
a & =-90.0\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

In terms of g taking absolute values of the acceleration , we have

\begin{align*}
a & = 90.0 \ \text{m/s}^2 \cdot \left( \frac{g}{9.80 \ \text{m/s}^2} \right) \\
a & = \frac{90.0}{9.80}g \\
a & = 9.18g \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We shall use the formula

\Delta x=v_{ave}t

where v_{ave} is the average velocity computed as 

v_{ave}=\frac{v_0+v_f}{2}

Solving for time t in terms of the other variables, we have

t=\frac{2\Delta x}{v_0+v_f}

Substituting the given values, we have

\begin{align*}
t & =\frac{2\Delta x}{v_0+v_f} \\
t & =\frac{2\left(0.002\:\text{m}\right)}{0.600\:\text{m/s}+0.000\:\text{m/s}} \\
t & =6.67\times 10^{-3}\:\text{s}  \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

Employing the same formula we used in Part A, we have

\begin{align*}
a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\
a & =\frac{\left(0.000\:\text{m/s}\right)^2-\left(0.600\:\text{m/s}\right)^2}{2\left(0.0045\:\text{m}\right)} \\
a & =-40.0\:\text{m/s}^2 
\end{align*}

In terms of g, taking absolute values of the acceleration

\begin{align*}
a & = 40.0 \ \text{m/s}^2 \cdot \left( \frac{g}{9.80 \ \text{m/s}^2} \right) \\
a & =\frac{40.0}{9.80}g \\
a & = 4.08g \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Mechanics of Materials: An Integrated Learning Approach 3rd Edition by Timothy A. Philpot Problem P1.2

A 2024-T4 aluminum tube with an outside diameter of 2.50 in. will be used to support a 27-kip load. If the axial normal stress in the member must be limited to 18 ksi, determine the wall thickness required for the tube.

Solution:

We are given the following values:

\begin{align*}
\text{Outside Diameter}, D & = 2.50\ \text{in} \\
\text{Axial Load}, P & = 27\ \text{kips} \\
\text{Maximum Axial Stress}, \sigma & = 18\ \text{ksi}\\
\text{Inside Diameter}, d & = D-2t
\end{align*}

We are solving for the unknown wall thickness of the tube, t.

From the definition of normal stress, solve for the minimum area required to support a 27-kip load without exceeding a stress of 18 ksi

\begin{align*}
\sigma & =\frac{P}{A} \\ 
A_{\text{min}} & =\frac{P}{\sigma } \\ 
A_{\text{min}} & = \frac{27\:\text{kips}}{18\:\text{ksi}} \\ 
A_{\text{min}} & = 1.500\:\text{in}^2 
\end{align*}

The cross-sectional area of the aluminum tube is given by

A=\frac{\pi }{4}\left(D^2-d^2\right)

Set this expression equal to the minimum area and solve for the maximum inside diameter, d

\begin{align*}
A & =\frac{\pi }{4}\left(D^2-d^2\right) \\
A & = \frac{\pi }{4}\left[\left(2.50\ \text{in}\right)^2-d^2\right] \\
1.500\ \text{in}^2 & = \frac{\pi }{4}\left[\left(2.50\ \text{in}\right)^2-d^2\right] \\
4 \left( 1.500\ \text{in}^2 \right) & = \pi \left[\left(2.50\ \text{in}\right)^2-d^2\right] \\
\frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi} & = \left(2.50\ \text{in}\right)^2-d^2 \\
d^{2} & = \left( 2.50\ \text{in}^{2} \right) - \frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi} \\
d & = \sqrt{\left( 2.50\ \text{in}^{2} \right) - \frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi}} \\
d & = 2.08330\ \text{in}
\end{align*}

The outside diameter D, the inside diameter d, and the wall thickness t are related by

D = d+2t

Therefore, the minimum wall thickness required for the aluminum tube is

\begin{align*}
t & =\frac{D-d}{2} \\
t & = \frac{2.50\:\text{in}-2.08330\:\text{in}}{2} \\
t & = 0.208\ \text{in} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Mechanics of Materials: An Integrated Learning Approach 3rd Edition by Timothy Philpot Problem P1.1

A stainless steel tube with an outside diameter of 60 mm and a wall thickness of 5 mm is used as a compression member. If the axial normal stress in the member must be limited to 200 MPa, determine the maximum load P that the member can support.

Solution:

We are given the following values:

\begin{align*}
\text{Outside Diameter}, D &= 60\ \text{mm} \\
\text{Wall Thickness}, t & = 5\ \text{mm} \\
\text{Inside Diameter}, d & = D-2t = 60\ \text{mm}-2\left( 5\ \text{mm} \right) = 50\ \text{mm}\\
\text{Maximum Axial Stress}, \sigma & =200\ \text{MPa} = 200\ \frac{\text{N}}{\text{mm}^2}
\end{align*}

The cross-sectional area of the stainless-steel tube is

\begin{align*}
A & = \frac{\pi}{4}\left( D^2 - d^2 \right) \\
A & = \frac{\pi}{4}\left[ \left( 60\ \text{mm} \right)^2 - \left( 50\ \text{mm} \right)^2 \right] \\
A & = 863.938\ \text{mm}^2
\end{align*}

The normal stress in the tube can be expressed as

\sigma =\frac{P}{A}

The maximum normal stress in the tube must be limited to 200 MPa. Using 200 MPa as the allowable normal stress, rearrange this expression to solve for the maximum load P.

\begin{align*}
P_{max} & = \sigma _{max} A \\
P_{max} & = \left( 200\ \text{MPa} \right)\left( 863.938\ \text{mm}^2 \right)\\
P_{max} & = \left( 200\ \frac{\text{N}}{\text{mm}^2} \right)\left( 863.938\ \text{mm}^2 \right)\\
P_{max} & = 172788\ \text{N} \\
P_{max} & = 172.8\ \text{kN}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

College Physics by Openstax Chapter 2 Problem 31

A swan on a lake gets airborne by flapping its wings and running on top of the water.

(a) If the swan must reach a velocity of 6.00 m/s to take off and it accelerates  from rest at an average rate of  0.350 m/s2, how far will it travel before becoming airborne?

(b) How long does this take?

Solution:

Part A

We are given the following values: v_f=6.00\:\text{m/s}; v_0=0\:\text{m/s}; and a=0.350\:\text{m/s}^2.

From the kinematic equations, the most applicable formula to solve for the change in distance, \Delta \text{x}, is

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Solving for \Delta \text{x} in terms of the other variables, we have

\Delta x=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2a}

Substituting the given values, 

\begin{align*}
\Delta x & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2a} \\
\Delta x  & =\frac{\left(6.00\:\text{m/s}\right)^2-\left(0.00\:\text{m/s}\right)^2}{2\left(0.350\:\text{m/s}^2\right)} \\
\Delta x  & =51.4286\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

From the formula v_f=v_0+at, solve for time, t in terms of the other variables.

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

Substitute the given values

\begin{align*}
t & =\frac{v_f-v_0}{a} \\
t & =\frac{6.00\:\text{m/s}-0.00\:\text{m/s}}{0.350\:\text{m/s}^2} \\
t & =17.1429\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 2 Problem 30

A fireworks shell is accelerated from rest to a velocity of 65.0 m/s over a distance of 0.250 m.

(a) How long did the acceleration last?

(b) Calculate the acceleration.

Solution:

We are given the following values:v_0=0\:\text{m/s}; v_f=65.0\:\text{m/s}; and \Delta x=0.250\:\text{m}.

We can immediately solve for the acceleration using the given values, so we are going to answer Part B first.

Part B

Solve for the acceleration first using the formula

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

We solve for acceleration in terms of the other variables.

a=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substitute the given values

\begin{align*}
a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\
a & = \frac{\left(65.0\:\text{m/s}\right)^2-\left(0\:\text{m/s}\right)^2}{2\left(0.250\:\text{m}\right)} \\
a & =8450\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part A

To solve for the time of this motion, we shall use the formula

v_f=v_0+at

Solving for time, t, in terms of the other variables we have.

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

We now substitute the values given, and the computed acceleration to find the time.

\begin{align*}
t & =\frac{v_f-v_0}{a} \\
t & =\frac{65.0\:\text{m/s}-0\:\text{m/s}}{8450\:\text{m/s}^2} \\
t & =7.6922\:\times 10^{-3}\:\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 2 Problem 29

Freight trains can produce only relatively small accelerations and decelerations.

(a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s2 for 8.00 minutes, starting with an initial velocity of 4.00 m/s?

(b) If the train can slow down at a rate of 0.550 m/s2, how long will it take to come to a stop from this velocity?

(c) How far will it travel in each case?

Solution:

Part A

We are given the the following: a=0.0500 \ \text{m/s}^2; t=8.00 \ \text{mins}; and v_0=4.00 \ \text{m/s}.

The final velocity can be solved using the formula v_f=v_0+at. We substitute the given values.

\begin{align*}
v_f&  = v_0+at \\
v_f & = 4.00\:\text{m/s}+\left(0.0500\:\text{m/s}^2\right)\left(8.00\:\text{min}\times \frac{60\:\sec }{1\:\min }\right) \\
v_f & = 28.0 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

Rearrange the equation we used in part (a) by solving in terms of t, we have

\begin{align*}
t & =\frac{{v_f}-v_0}{a} \\
t & = \frac{0\:\text{m/s}-28\:\text{m/s}}{-0.550\:\text{m/s}^2} \\
t & = 50.91\:\sec\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

The change in position for part (a), \Delta x, or distance traveled is computed using the formula  \Delta x=v_0 t+\frac{1}{2} at^2.

\begin{align*}
 \Delta x & =v_0 t+\frac{1}{2} at^2 \\
\Delta x & =\left(4.0\:\text{m/s}\right)\left(480\:\text{s}\right)+\frac{1}{2}\left(0.0500\:\text{m/s}^2\right)\left(480\:\text{s}\right)^2 \\
 \Delta x & = 7680\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

For the situation in part (b), the distance traveled is computed using the formula \Delta x=\frac{v_f^2-v_0^2}{2 a}.

\begin{align*}
\Delta x & =\frac{\left(0\:\text{m/s}\right)^2-\left(28.0\:\text{m/s}\right)^2}{2\left(-0.550\:\text{m/s}^2\right)} \\
\Delta x  & =712.73\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 2 Problem 28

A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/h) in only 3.90 s.

(a) What is its average acceleration?

(b) How far does it travel in that time?

Solution:

We are given the following: v_0=0 \ \text{m/s}; v_f=26.8 \ \text{m/s}; and t=3.90\ \text{s}.

Part A

The average acceleration of the motorcycle can be solved using the equation \overline{a}=\frac{\Delta v}{\Delta t}. Substitute the given into the equation. That is,

\begin{align*}
\overline{a} & =\frac{\Delta v}{\Delta t} \\
\overline{a} & =\frac{26.8\:\text{m/s}-0\:\text{m/s}}{3.90\:\text{s}} \\
\overline{a} & =6.872\:\text{m/s}^2\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The distance traveled is equal to the average velocity multiplied by the time of travel. That is,

\begin{align*}
\Delta x & =v_{ave}t\\
\Delta x & =\left(\frac{0\:\text{m/s}+26.8\:\text{m/s}}{2}\right)\left(3.90\:\text{s}\right) \\
\Delta x & =52.26\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}
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College Physics by Openstax Chapter 2 Problem 27

In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If this shot takes 3.33×10-2 s, calculate the distance over which the puck accelerates.

Solution:

The best equation that can be used to solve this problem is \Delta x=v_{ave} t. That is,

\begin{align*}
\Delta x & = v_{ave} t \\
\Delta x & = \left(\frac{8\:\text{m/s}+40\:\text{m/s}}{2}\right)\left(3.33\times 10^{-2}\:\text{s}\right) \\
\Delta x & = 0.7992\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the distance over which the puck accelerates is 0.7992 meters.

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College Physics by Openstax Chapter 2 Problem 26

Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm by the left ventricle of the heart.

(a) Make a sketch of the solution.

(b) List the knowns in this problem.

(c) How long does the acceleration take? To solve this part, identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units.

(d) Is the answer reasonable when compared with the time for a heartbeat?

Solution:

Part A

The sketch should contain the starting point and the final point. This will be done by connecting a straight line from the starting point to the final point. The sketch is shown below.

Part B

The list of known variables are:

Initial velocity: v_0=0\:\text{m/s}
Final Velocity: v_f=30.0\:\text{cm/s}
Distance Traveled: x-x_0=1.80\:\text{cm}

Part C

The best equation to solve for this is \Delta \text{x}=\text{v}_{\text{ave}}\text{t} where v_{ave} is the average velocity, and t is time. That is

\begin{align*}
\Delta x & =v_{ave} t \\
t &=\frac{\Delta x}{v_{ave}} \\
t & =\frac{1.80\:\text{cm}}{\frac{\left(0\:\text{cm/s}+30\:\text{cm/s}\right)}{2}}\\
t & =0.12\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Part D

Since the computed value of the time for the acceleration of blood out of the ventricle is only 0.12 seconds (only a fraction of a second), the answer seems reasonable. This is due to the fact that an entire heartbeat cycle takes about one second. So, the answer is yes, the answer is reasonable.

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College Physics by Openstax Chapter 2 Problem 25

At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of 2.00 m/s2.

(a) How far does she travel in the next 5.00 s?

(b) What is her final velocity?

(c) Evaluate the result. Does it make sense?

Solution:

We are given the following: v_0=9.00 \ \text{m/s}; and a=2.00 \ \text{m/s}^2.

Part A

For this part, we are given t=5.00 \ \text{s} and we shall use the formula  x=x_0+v_0 t+\frac{1}{2}at^2.

\begin{align*}
x & =x_0+v_0 t+\frac{1}{2}at^2 \\
x & =0\:\text{m}+\left(9.00\:\text{m/s}\right)\left(5.00\:\text{s}\right)+\frac{1}{2}\left(-2.00\:\text{m/s}^2\right)\left(5.00\:\text{s}\right)^2 \\
x & =20\:\text{meters} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

Part B

The final velocity can be determined using the formula v_f=v_0+at.

\begin{align*}
v_f & =v_0+at \\
v_f & =9.00\:\text{m/s}+\left(-2.00\:\text{m/s}^2\right)\left(5.00\:\text{s}\right) \\
v_f & =-1\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

Part C

The result says that the runner starts at the rate of 9 m/s and decelerates at 2 m/s2. After some time, the velocity is already negative. This does not make sense because if the velocity is negative, that means that the runner is already running backwards.

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