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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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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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College Physics by Openstax Chapter 2 Problem 24

While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s² for 12.0 s.

(a) Draw a sketch of the situation.

(b) List the knowns in this problem.

(c) How far does the car travel in those 12.0 s? To solve this part, first 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, check your units, and discuss whether the answer is reasonable.

(d) What is the car’s final velocity? Solve for this unknown in the same manner as in part (c), showing all steps explicitly.

Solution:

Part A

The sketch of the situation is shown below. Also, the knowns and unknowns are in the illustration.

College Physics Problem 2.24 Illustration

From the illustration above, we can see that the initial velocity is 0 m/s, the initial time and initial distance are also zero. The final velocity and the final distance are unknowns. The time at the final location is 12 seconds and the acceleration is constant all throughout the trip at 2.40 meters per second square.

Part B

The knowns are: $a=2.40\:\text{m/s}^2$ ; $t=12.0\:\sec$ ; $v_0=0\:\text{m/s}$ ; and $x_0=0\:\text{m}$

Part C

For this part, the unknown is the value of x. If we examine the equations for constant acceleration and the given values in this problem, we can readily use the formula $x=x_0+v_0 t+\frac{1}{2}at ^2$ . That is

\begin{align*} x & =x_0+v_0 t+\frac{1}{2}at^2 \\ x & =0\:\text{m}+\left(0\:\text{m/s}\right)\left(12.0\:\text{s}\right)+\frac{1}{2}\left(2.40\:\text{m/s}^2\right)\left(12.0\:\text{s}\right)^2 \\ x & =172.8\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}

Part D

For this part, the unknown is the value of v. The equation that can be used based from the known variables is v=v_0+at. That is

\begin{align*} v & =v_0+at \\ v & =0\:\text{m/s}+\left(2.40\:\text{m/s}^2\right)\left(12.0\:\text{s}\right) \\ v & =28.8\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics by Openstax Chapter 2 Problem 23

a) A light-rail commuter train accelerates at a rate of 1.35 m/s². How long does it take to reach its top speed of 80.0 km/h, starting from rest?

b) The same train ordinarily decelerates at a rate of 1.65 m/s². How long does it take to come to a stop from its top speed?

c) In emergencies, the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in m/s²?

Solution:

Part A

We are given the following: $a=1.35 \ \text{m/s}^2$ ; $v_f=80.0 \ \text{km/h}$ ; and $v_0=0 \ \text{m/s}$ .

From the formula $v_f=v_0+at$ , we can solve for $t$ as

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

We need to convert 80.0 km/h to m/s first so that we have a unit uniformity for all the given values.

\begin{align*} 80\:\text{km/hr} & =\left(80\:\text{km/hr}\right)\left(\frac{1000\:\text{m}}{1\:\text{km}}\right)\left(\frac{1\:\text{hr}}{3600\:\text{s}}\right) \\ & =22.2222\:\text{m/s} \end{align*}

Substituting the given values into the formula, we have

\begin{align*} t & =\frac{v_f-v_0}{a} \\ t & =\frac{22.2222\:\text{m/s}-0\:\text{m/s}}{1.35\:\text{m/s}^2} \\ t & =16.5\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

For this problem, we still use the formula used in Part (a). This time, the values of the final and initial velocities interchange and the value of the given deceleration is negative of acceleration. The given values are $a=-1.65 \ \text{m/s}^2$ ; v_0=22.2222 \ \text{m/s}[/katex]; and $v_f=0 \ \text{m/s}$

\begin{align*} t & =\frac{v_f-v_0}{a} \\ t & =\frac{0\:\text{m/s}-22.2222\:\text{m/s}}{-1.65\:\text{m/s}^2} \\ t & =13.5\:\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

For this problem, we will use the formula

\begin{align*} a & =\frac{v_f-v_0}{\Delta t} \end{align*}

Substituting all the given values into the formula, we have

\begin{align*} a & =\frac{0\:\text{m/s}-22.2222\:\text{m/s}}{8.30\:\text{s}} \\ a & =2.68\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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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?

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?

Part B

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.

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?

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?

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.

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?

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?

While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s.

(a) Draw a sketch of the situation.

(b) List the knowns in this problem.

(d) What is the car’s final velocity? Solve for this unknown in the same manner as in part (c), showing all steps explicitly.

a) A light-rail commuter train accelerates at a rate of 1.35 m/s2. How long does it take to reach its top speed of 80.0 km/h, starting from rest?

b) The same train ordinarily decelerates at a rate of 1.65 m/s2. How long does it take to come to a stop from its top speed?

c) In emergencies, the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in m/s2?

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

(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?

(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?

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.

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.

While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s² for 12.0 s.

a) A light-rail commuter train accelerates at a rate of 1.35 m/s². How long does it take to reach its top speed of 80.0 km/h, starting from rest?

b) The same train ordinarily decelerates at a rate of 1.65 m/s². How long does it take to come to a stop from its top speed?

c) In emergencies, the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in m/s²?