Tag Archives: Physics Solutions

College Physics by Openstax Chapter 2 Problem 60

Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t=30.0 s is approximately 0.24 m/s. Assume all values are known to 2 significant figures.

Figure 2.62

Solution:

We can obviously see from the graph that it is a straight line or approximately a straight line. In this case, the slope is constant.

To get an approximate slope at t=30 s, we can use the values at t=20 s and t=40 s. When t=20\ \text{s}, x=7\ \text{m} and when t=40\ \text{s}, x=12\ \text{m}.

\begin{align*}

\text{slope} & =\frac{\Delta x}{\Delta t} \\
\text{slope} & = \frac{x_2-x_1}{t_2-t_1} \\
\text{slope} & = \frac{12\ \text{m}-7\ \text{m}}{40\ \text{s}-20\ \text{s}} \\
\text{slope} & =0.25\ \text{m/s}

\end{align*}

Although not equal, the computed slope is almost the same with 0.24 m/s. This is due to the fact that values are uncertain when using graphs. The difference is not really significant for this case.

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College Physics by Openstax Chapter 2 Problem 59

(a) By taking the slope of the curve in Figure 2.60, verify that the velocity of the jet car is 115 m/s at t=20 s. (b) By taking the slope of the curve at any point in Figure 2.61, verify that the jet car’s acceleration is 5.0 m/s2 .

Figure 2.60
Figure 2.61

Solution:

Part A

Figure A

Figure A shows the approximate slope of the curve at time 20 seconds.

To solve for the slope of this line, we need to approximate by using two points. In this case, we shall use the points at time 15 seconds and 25 seconds.

Approximately, when t=15\ \text{s}, the position is x=1000\ \text{m}, and when t=25\ \text{s}, the position is x=2150\ \text{m}. Thefore,

\begin{align*}

\text{velocity }& = \text{slope} \\
v& = \frac{\Delta x}{\Delta t} \\
v& = \frac{x_2-x_1}{t_2-t_1} \\
v& = \frac{2150\ \text{m}-1000\ \text{m}}{25\ \text{s}-15\ \text{s}}\\
v& = 115\ \text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}

\end{align*}

Part B

One can immediately figure out from the given graph that it is a straight line. The slope of the line can be computed by using any two points in the line.

Here, v=15\ \text{m/s} when t=0\ \text{s}, and v=40 \ \text{m/s} when t=5\ \text{s}. The acceleration is

\begin{align*}

\text{acceleration}& = \text{slope} \\
a& = \frac{\Delta v}{\Delta t}\\
a& = \frac{v_2-v_1}{t_2-t_1} \\
a& = \frac{40\ \text{m/s}-15\ \text{m/s}}{5\ \text{s}-0\ \text{s}}\\
a& = 5.0\ \text{m/s}^2 \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}

\end{align*}
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College Physics by Openstax Chapter 2 Problem 58

A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms (3.50×10−3s) . (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

Solution:

The concept is the same with Problem 2.56.

Part A

\begin{align*}

\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\

\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1  \right) \\

 v_{y_2} & = \sqrt{\left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1  \right)} \\

 v_{y_2} & = -  \sqrt{\left( 0 \right)^2+2\left( -9.81\ \text{m/s}^2 \right)\left( 0\ \text{m} - 1.50\ \text{m} \right)} \\

 v_{y_2} & = - 5.42 \ \text{m/s} \qquad {\color{DarkOrange} \left( \text{Answer} \right)}



\end{align*}

Part B

\begin{align*}

\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\

\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1  \right) \\

\left( v_{y_1} \right)^2 & = \left( v_{y_2} \right)^2 - 2 a \left( y_2 - y_1  \right) \\

v_{y_1} & = \sqrt{\left( v_{y_2} \right)^2 - 2 a \left( y_2 - y_1  \right)} \\

v_{y_1} & = \sqrt{\left( 0 \ \text{m/s} \right)^2 -2\left( -9.81 \ \text{m/s}^2  \right) \left( 1.10 \ \text{m} - 0 \ \text{m}  \right) }\\

v_{y_1} & =  4.65 \ \text{m/s}  \qquad {\color{DarkOrange} \left( \text{Answer} \right)}

\end{align*}

Part C

\begin{align*}

a & = \frac{\Delta v}{\Delta t} \\
a & = \frac{v_2-v_1}{\Delta t} \\
a & = \frac{4.65 \ \text{m/s} - \left( -5.42 \ \text{m/s} \right)}{3.50 \times 10^{-3} \ \text{s}} \\
a & = 2877 \ \text{m/s}^2 \\
a & = 2.88 \times 10^{3}\    \text{m/s}^2 \ \qquad  {\color{DarkOrange} \left( \text{Answer} \right)}\\

\end{align*}

Part D

\begin{align*}
\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\
\Delta y & =  \frac{\left( v_{y_2} \right)^2 -  \left( v_{y_1} \right)^2}{2 a} \\
\Delta y & = \frac{\left( 0 \ \text{m/s} \right)^2 - \left( -5.42 \ \text{m/s} \right)^2}{2\left( 2.88 \times 10^3 \ \text{m/s}^2 \right)}\\
\Delta y & = -0.00510 \ \text{m} \\
 \Delta y & = -5.10 \times 10^{-3} \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}
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College Physics by Openstax Chapter 2 Problem 57

A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at 10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s after being released, and (c) the time before it hits the ground.

Solution:

Part A

Figure A

Consider Figure A.

We are interested in two positions. Position 1 is where the coin is dropped. At this position, the coin is 300 m above the ground, the time is 0 s, and the velocity is 10.0 m/s upward.

Position 2 is the highest point of the coin reaches. At this position, the velocity is equal to 0 m/s.

Position 1 is the initial position and position 2 is the final position. Solve for the value of y2.

\begin{align*}
\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 +2a \Delta y \\
\Delta y & = \frac{\left( v_{y_2} \right)^2 - \left( v_{y_1} \right)^2}{2a} \\
y_2 - y_1 & =  \frac{\left( v_{y_2} \right)^2 - \left( v_{y_1} \right)^2}{2a} \\
y_2& =  \frac{\left( v_{y_2} \right)^2 - \left( v_{y_1} \right)^2}{2a} +y_1 \\
y_2 & = \frac{\left( 0 \ \text{m/s} \right)^2-\left( 10.0\ \text{m/s} \right)^2}{2\left( -9.81 \ \text{m/s}^2 \right)}+300\ \text{m}
\\
y_2 & =305 \ \text{m} \ \qquad  {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}

\therefore The maximum height reached by the coin is about 305 meters from the ground.

Part B

We do not know the position 4 seconds after the coin has been released, the answer can be above or below the initial point. We can actually use one of the kinematical equations to solve for the final position given the time. Here, the initial position is the point of release and the final position is the point of interest at 4.00 seconds after release.

\begin{align*}
\Delta y & = v_{y_1}t+\frac{1}{2}at^2 \\
y_2 - y_1 & = v_{y_1}t+\frac{1}{2}at^2 \\
y_2 & = y_1 +  v_{y_1}t+\frac{1}{2}at^2 \\
y_2 & = 300\ \text{m} +\left( 10\ \text{m/s} \right)\left( 4.00\ \text{s} \right)+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( 4.00\ \text{s} \right)^2\\
y_2 & = 261.52\ \text{m} \\
y_2 & = 262\ \text{m}\ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}  \\
\end{align*}

\therefore The coin is at a height of 262 meters above the ground 4.00 seconds after release. That is, the coin is already dropping and it is already below the release point.

Solving for the velocity 4.00 seconds after release considering the same initial and final position.

\begin{align*}
v_{y_2} & = v_{y_1}+at \\
v_{y_2} & = 10\ \text{m/s} + \left( -9.81\ \text{m/s}^2 \right)\left( 4.00 \ \text{s} \right) \\
v_{y_2} & = -29.2\ \text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

\therefore The coin has a velocity of 29.2 m/s directed downward 4.00 seconds after it is released. This confirms that the coin is indeed moving downwards at this point.

Part C

Figure C

Considering figure C, we have two positions. Position 1 is the point of release 300 m above the ground with a velocity of 10 m/s upward. This is time 0 s.

The second position is at the ground where y=0 m. We are interested at the time in this position.

Considering position 1 as the initial position and position 2 as the final position.

\begin{align*}
\Delta y & = v_{y_1} \Delta t+\frac{1}{2}a\left( \Delta t \right)^2 \\
y_2-y_1 & =  v_{y_1}t+\frac{1}{2}at^2 \\
0\ \text{m}-300\ \text{m} & = \left( 10 \ \text{m/s} \right)t+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)t^2 \\
-300 & = 10t-4.905t^2 \\
4.905t^2-10t-300 & = 0 \\
\end{align*}

Solve for the value of t using the quadratic formula with a=4.905, b=-10, and c=-300.

\begin{align*}
t & = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\\
t & = \frac{-\left( -10 \right) \pm \sqrt{\left( 10 \right)^2-4\left( 4.905 \right)\left( -300 \right)}}{2\left( 4.905 \right)}\\
t & = 8.91 \ \text{s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

\therefore The time is about 8.91 seconds before the coin hits the ground.

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College Physics by Openstax Chapter 2 Problem 56

A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms (8.00×10−5 s) . (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

Solution:

Part A

Figure A

For this part, we shall consider Figure A.

We will be considering the two positions as shown. The first position is when the ball is dropped from a height of 1.50 meters. For this position, we know that y1=1.50 m, t1=0 s, and vy1=0 m/s.

Position 2 is immediately after the ball hits the floor. For this position, we do not the time elapse and the velocity but we know that the height is zero. That is y2=0 m.

Position 1 is the initial position and position 2 is the final position. Solving for vy2, we have

\begin{align*}

\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\

\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1  \right) \\

 v_{y_2} & = \sqrt{\left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1  \right)} \\

 v_{y_2} & = -  \sqrt{\left( 0 \right)^2+2\left( -9.81\ \text{m/s}^2 \right)\left( 0\ \text{m} - 1.50\ \text{m} \right)} \\

 v_{y_2} & = - 5.42 \ \text{m/s} \qquad {\color{DarkOrange} \left( \text{Answer} \right)}



\end{align*}

\therefore The steel ball has a velocity of about 5.42 m/s directed downward when it strikes the floor.

Part B

Figure B

Figure B shows the two positions we are interested in to solve for this part.

Position 1 is at the floor immediately just after the ball hits it. At this initial position, we have y1=0 m, and t1= 0 s. We do not know the velocity at this point.

At position 2, the ball bounced back to its second peak. We know that at the peak of a free falling body, the velocity is zero. So, for this final position, we have y2=1.45 m, and vy2=0 m/s. We do not know the time at this position.

Position 1 is the initial position while position 2 is the final position. Solving for the initial velocity we have

\begin{align*}

\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\

\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1  \right) \\

\left( v_{y_1} \right)^2 & = \left( v_{y_2} \right)^2 - 2 a \left( y_2 - y_1  \right) \\

v_{y_1} & = \sqrt{\left( v_{y_2} \right)^2 - 2 a \left( y_2 - y_1  \right)} \\

v_{y_1} & = \sqrt{\left( 0 \ \text{m/s} \right)^2 -2\left( -9.81 \ \text{m/s}^2  \right) \left( 1.45 \ \text{m} - 0 \ \text{m}  \right) }\\

v_{y_1} & =  5.33 \ \text{m/s}  \qquad {\color{DarkOrange} \left( \text{Answer} \right)}

\end{align*}

\therefore The steel ball has a velocity of about 5.33 m/s directed upward immediately after it leaves the floor.

Part C

From our answers in Part A and Part B, we have a change in velocity from -5.42 m/s to 5.33 m/s. So, in this case the initial velocity is v1=-5.42 m/s and the final velocity is v2=5.33 m/s. We can compute for the acceleration:

\begin{align*}

a & = \frac{\Delta v}{\Delta t} \\
a & = \frac{v_2-v_1}{\Delta t} \\
a & = \frac{5.33 \ \text{m/s} - \left( -5.42 \ \text{m/s} \right)}{8.00 \times 10^{-5} \ \text{s}} \\
a & = 134,375 \ \text{m/s}^2 \\
a & = 1.34 \times 10^{5}   \text{m/s}^2 \ \qquad  {\color{DarkOrange} \left( \text{Answer} \right)}\\

\end{align*}

Part D

The period of compression happens when the ball has a velocity of -5.42 m/s until it reaches 0 m/s. We shall solve for the change in displacement for this two given velocities. The initial velocity is -5.42 m/s and the final velocity is 0 m/s. The acceleration during this period is the one solved in Part C, a=1.34×105 m/s2. 

\begin{align*}
\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\
\Delta y & =  \frac{\left( v_{y_2} \right)^2 -  \left( v_{y_1} \right)^2}{2 a} \\
\Delta y & = \frac{\left( 0 \ \text{m/s} \right)^2 - \left( -5.42 \ \text{m/s} \right)^2}{2\left( 1.34 \times 10^5 \ \text{m/s}^2 \right)}\\
\Delta y & = -0.000110 \ \text{m} \\
 \Delta y & = -1.10 \times 10^{-4} \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}
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College Physics by Openstax Chapter 2 Problem 55

Suppose you drop a rock into a dark well and, using precision equipment, you measure the time for the sound of a splash to return.
(a) Neglecting the time required for sound to travel up the well, calculate the distance to the water if the sound returns in 2.0000 s.
(b) Now calculate the distance taking into account the time for sound to travel up the well. The speed of sound is 332.00 m/s in this well.

Solution:

Part A

Figure A

Consider Figure A.

We shall consider two points for our solution. First, position 1 is the top of the well. In this position, we know that y1=0, t1=0 and vy1=0.

Position 2 is located at the top of the water table where the rock will meet the water. Since we neglect the time for the sound to travel from position 2 to position 1, we can say that t2=2.0000 s, the time of the rock to reach this position.

Solving for the value of y2 will determine the distance between the two positions.

\begin{align*}

\Delta y & = v_{y_1}t+\frac{1}{2}at^2  \\
y_2-y_1 & = v_{y_1}t+\frac{1}{2}at^2  \\
y_2 & = y_1 +v_{y_1}t+\frac{1}{2}at^2  \\
y_2 & = 0+0+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( 2.0000\ \text{s} \right)^2 \\
y_2 & = -19.6\ \text{m} \qquad {\color{DarkOrange} \left( \text{Answer }\right)}

\end{align*}

Position 2 is 19.6 meters measured downward from position 1.

\therefore The distance to the water is about 19.6 meters.

Part B

For this case, the 2.0000 seconds that is given includes the time that the rock travels from position 1 to position 2, tr, and the time that the sound travels from position 2 to position 1, ts.

\begin{align*}
t_r+t_s & =2.0000\ \text{s} \\
t_s & = 2.0000\ \text{s}-t_r 
\end{align*}

Considering the motion of the rock from position 1 to position 2.

\begin{align*}

\Delta y & = v_{y_1}t_r+\frac{1}{2}a\left( t_r \right)^2  \\
y_2-y_1 & = v_{y_1}t_r+\frac{1}{2}a\left( t_r \right)^2  \\
y_2 & = y_1 +v_{y_1}t_r+\frac{1}{2}a\left( t_r \right)^2  \\
y_2 & = 0+0+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( t_r \right)^2  \\
y_2 & = -4.905\left( t_r \right)^2  \qquad {\color{Blue} \text{Equation 1}}\\


\end{align*}

Now, let us consider the motion of the sound from position 2 to position 1. Sound is assumed to have a constant velocity of 322.00 m/s.

\begin{align*}

\Delta y & = v_s \times t_s \\
y_1-y_2 & =\left( 322.00\ \text{m/s} \right)\left( t_s \right) \\
0-y_2 & =\left( 322.00 \right)\left( 2.0000-t_r \right) \\
y_2 & = -322.00\left( 2.0000-t_r \right) \qquad  {\color{Blue} \text{Equation 2}}


\end{align*}

So, we have two equations from the two motions. We can solve the equations simultaneously. 

\begin{align*}

-4.905 \left( t_r \right)^2 & = -322.00\left( 2.0000-t_r \right) \\
4.905 \left( t_r \right)^2 & =322.00\left( 2.0000-t_r \right) \\
4.905 \left( t_r \right)^2 & = 644.00-322.00t_r \\
4.905\left( t_r \right)^2 + 322.00t_r-644.00 & = 0 \\

\end{align*}

We can solve the quadratic formula using the quadratic equation. 

\begin{align*}

t_r & = \frac{-b \pm\sqrt{b^2-4ac}}{2a} \\
t_r & = \frac{-322.00\pm\sqrt{\left( 322.00 \right)^2-4\left( 4.905 \right)\left( -644.00 \right)}}{2\left( 4.905 \right)}\\
t_r & =1.9425 \ \text{s}

\end{align*}

Now that we have solved for the value of tr, we can use this to solve for y2 using either Equation 1 or Equation 2. We will use equation 1.

\begin{align*}

y_2 & = -4.905\left( t_r \right)^2 \\
y_2 & = -4.905 \left( 1.9425 \right)^2 \\
y_2 & =-18.5 \ \text{m} \qquad {\color{DarkOrange} \left( \text{Answer} \right)}

\end{align*}

Position 2 is about 18.5 meters below position 1.

\therefore In this case, the distance between the two positions is 18.5 meters.

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College Physics by Openstax Chapter 2 Problem 54

A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 0.312 s to go past the window. What was the ball’s initial velocity? Hint: First consider only the distance along the window, and solve for the ball’s velocity at the bottom of the window. Next, consider only the distance from the ground to the bottom of the window, and solve for the initial velocity using the velocity at the bottom of the window as the final velocity.

Solution:

First, we have the position 1 where the motion starts. Here, we know that y<sub>1</sub>=0, t<sub>1</sub>=0, and v<sub>y1</sub>=0. Position 2 is at the bottom of the window. We know that it is 7.50 meters from where the motion started. So we have y<sub>2</sub>=7.50 meters. We do not know the time and velocity at this point. Then we have position 3 at the top of the window where the overall height is 9.50 meters, y<sub>3</sub>=9.50. We also do not know the velocity and time elapsed in this position.
Figure A

Consider Figure A. We shall be considering the three positions shown.

First, we have position 1 where the motion starts. Here, we know that y1=0 and t1=0, but we do not know vy1.

Position 2 is at the bottom of the window. We know that it is 7.50 meters from where the motion started. So we have y2=7.50 meters. We do not know the time and velocity at this point.

Then we have position 3 at the top of the window where the overall height is 9.50 meters, y3=9.50. We also do not know the velocity and time elapsed in this position.

Consider positions 2 and 3. The initial position in this case is at position 2 and the final position is at position 3. We know that the difference of time between this two positions is 0.312 seconds. We can say that

t_3 =t_2+0.312 \ \text{s} \\
t_3-t_2 = 0.312\ \text{s}

Using the same 2 positions still, we have

\begin{align*}

y_3 & = y_2 + v_{y_2} \Delta t+\frac{1}{2}a\left(  \Delta t \right)^2 \\
9.50\ \text{m} & = 7.50\ \text{m} +  v_{y_2} \left( t_3-t_2 \right)+\frac{1}{2}a\left( t_3-t_2 \right)^2 \\
9.50\ \text{m}-7.50\ \text{m} & = v_{y_2}\left( 0.312\ \text{s} \right)+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( 0.312\ \text{s} \right)^2\\
2.00\ \text{m} & = 0.312\ \text{s} \left( v_{y_2} \right)-0.4775\ \text{m} \\
 0.312\ \text{s} \left( v_{y_2} \right) & = 2.00\ \text{m}+0.4775\ \text{m} \\
 0.312\ \text{s} \left( v_{y_2} \right) & = 2.4775\ \text{m} \\
v_{y_2}& =\frac{2.4775\ \text{m}}{0.312\ \text{s}} \\
v_{y_2}& = 7.94\ \text{m/s}

\end{align*}

We have computed the velocity of the ball at the bottom of the window.

Next, we shall consider positions 1 and 2. In this consideration, position 1 will be considered the initial position while position2 is the final position. 

\begin{align*}

\left( v_{y_2} \right)^2  &  = \left( v_{y_1} \right)^2 +2a \Delta y \\
\left( 7.94\ \text{m/s} \right)^2 & = \left( v_{y_1} \right)^2 + 2\left( -9.81 \ \text{m/s}^2 \right)\left( 7.50\ \text{m}-0 \right)\\
\left( v_{y_1} \right)^2 & = \left( 7.94\ \text{m/s} \right)^2- 2\left( -9.81 \ \text{m/s}^2 \right)\left( 7.50\ \text{m}-0 \right)\\
v_{y_1} & = + \sqrt{ \left( 7.94\ \text{m/s} \right)^2- 2\left( -9.81 \ \text{m/s}^2 \right)\left( 7.50\ \text{m}-0 \right)} \\
v_{y_1} & = + 14.5\ \text{m/s}\qquad {\color{DarkOrange} \left( \text{Answer} \right) }

\end{align*}

\therefore The ball’s initial velocity is about 14.5 m/s upward.

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College Physics by Openstax Chapter 2 Problem 53

There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff.
(a) How fast will it be going when it strikes the ground?
(b) Assuming a reaction time of 0.300 s, how long will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? The speed of sound is 335 m/s on this day.

Solution:

Part A

Consider the Figure A. We shall consider two position points — position 1 at the top of the cliff, and position 2 on the ground.

Position 1 is 250 meters above the ground (y=250 m), and since this is the initial position, t=0, and the initial velocity is vy=0.

Position 2 is on the ground (y=0), and we do not know the time and velocity at this point.

For this part, we will solve for the value of vy at position 2. 

\begin{align*}
\left( v_{y_{2}} \right)^2 &  = \left( v_{y_{1}} \right)^2+2a \Delta y \\
\left( v_{y_{2}} \right)^2 &  = \left( 0\ \text{m/s} \right)^2+2\left( -9.81\ \text{m/s}^2 \right)\left( 0\ \text{m}-250\ \text{m} \right) \\
\left( v_{y_{2}} \right)^2 &  = 4905\ \text{m}^2/\text{s}^2 \\
v_{y_{2}} & = \pm \sqrt{4905\ \text{m}^2/\text{s}^2} \\
v_{y_{2}} & = \pm \ 70.0\ \text{m/s}\\
v_{y_{2}} & = -  70.0\ \text{m/s} \qquad {\color{gold}\left( \text{Answer} \right) }\\
\end{align*}
Figure A

Since the boulder is moving downward at this point, we shall consider the negative sign of the velocity as an indication of the downward direction of the velocity.

To answer the question, the boulder is going at about 70.0 m/s downward when it strikes the ground.

Part B

Let:
t_s be the time for the sound to travel from the top of the cliff to the tourist;
t be the total time elapsed before the tourist can react;
t_b be the time for the rock to travel from the top of the cliff to the ground
t_t be the amount of time the tourist has to get out of the way after hearing the sound of the rock breaking loose

Since we are given the speed of sound, we have

\begin{align*}

t_s & = \frac{\text{height of the cliff}}{\text{speed of sound}} \\
t_s & = \frac{250\ \text{m}}{335\ \text{m/s}}\\
t_s & =0.746 \ \text{s}

\end{align*}

So, the tourist can react after

\begin{align*}

t & =t_s + \text{reaction time} \\
t & = 0.746\ \text{s}+0.300\ \text{s} \\
t &= 1.046\ \text{s}

\end{align*}

Then, we can compute for the total time it takes for the rock to travel from top to the ground. 

\begin{align*}

t_b & =\frac{v_{y_2}-v_{y_1}}{a} \\
t_b & =\frac{-70\ \text{m/s}-0\ \text{m/s}}{-9.81\ \text{m/s}^2} \\
t_b & = 7.14\ \text{s}

\end{align*}

Therefore, the tourist still has the time to get out of the way. That is,

\begin{align*}

t_t & =7.14\ \text{s}-1.046\ \text{s} \\
t_t & = 6.09\ \text{s} \qquad {\color{gold}\left( \text{Answer} \right) } \\

\end{align*}

\therefore the tourist has about 6.09 seconds to get out of the way.

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College Physics by Openstax Chapter 2 Problem 52

An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.

Solution:

Consider Figure 1. The object was dropped from a height of 75.0 m. At the start of motion, the velocity is zero, v_{oy}=0.

The object traveled for a period of time t for the whole 75.0 m distance to the ground.

Part A

We are solving for the distance traveled by the object for the first 1 second. So, we have

\begin{align*}
\Delta y & = v_{oy} t + \frac{1}{2}at^2\\
\Delta y & = \left( 0 \ \text{m/s} \right)\left( 1 \ \text{s} \right)+\frac{1}{2}\left( -9.81 \ \text{m/s}^2 \right)\left( 1 \ \text{s} \right)^2 \\
\Delta y & = 0 -4.905 \ \text{m} \\
\Delta y & = -4.91 \ \text{m} \\
|\Delta y| & =4.91 \ \text{m}
\end{align*}

The negative sign of Δy indicates that the direction of the displacement is downward. Since we are looking for the scalar value of the distance, the answer is 4.91 m.

Part B

So we now consider the two positions of the object as shown in the figure to the right. The initial height of the object is 75.0 m above the ground, and the initial velocity is 0.

At the ground, we know that the position of the object is 0 m above the ground, but we do not know the time and velocity. Therefore, to determine the velocity of the object at this point, we proceed as follows:

\begin{align*}
\left( v_2 \right)^2 & =\left( v_1 \right)^2+2a \Delta y \\
\left( v_2 \right)^2 & =\left( v_1 \right)^2+2a \left( y_2-y_1 \right) \\
v_2 & = \pm \sqrt{\left( v_1 \right)^2+2a \left( y_2-y_1 \right)}\\
v_2 & = \pm \sqrt{\left( 0\ \text{m/s} \right)^2+2\left( -9.81\ \text{m/s}^2 \right)\left( 0 \ \text{m}-75\ \text{m} \right)} \\
v_2 & = \pm \ 38.4\ \text{m/s}\\
v_2 & =- 38.4\ \text{m/s}\\
\end{align*}

Since the object is directing downwards when it hit the ground, the velocity is negative.

Part C

First, we calculate the total time of the object’s motion from the beginning to the ground. 

\begin{align*}
\Delta y & =\bcancel{v_{oy}t}+ \frac{1}{2}at^2 \\
\Delta y & = \frac{1}{2}at^2 \\
0 \text{m}-75\ \text{m} & = \frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)t^2 \\
t^2& =\frac{-75\ \text{m}}{-4.905 \text{m/s}^2}\\
t&=\sqrt{\frac{75\ \text{m}}{4.905 \text{m/s}^2}}\\
t&=3.91 \ \text{s}
\end{align*}

Second, determine the total distance traveled from 0 s to 2.91 s, leaving out the last 1 s of the motion. 

\begin{align*}
\Delta y & = v_{oy} t + \frac{1}{2}at^2\\
\Delta y & = \left( 0 \ \text{m/s} \right)\left( 1 \ \text{s} \right)+\frac{1}{2}\left( -9.81 \ \text{m/s}^2 \right)\left( 2.91 \ \text{s} \right)^2 \\
\Delta y & = 0 -41.5 \ \text{m} \\
\Delta y & = -41.5 \ \text{m} \\
|\Delta y| & =41.5 \ \text{m}
\end{align*}

Finally, subtract this distance from the total distance traveled to get the distance traveled in the last 1 second.

\begin{align*}
y_{_{\text{last 1 sec}}} & = 75.0 \ \text{m}-41.5 \ \text{m} \\
y_{_{\text{last 1 sec}}} & = 33.5\ \text{m}
\end{align*}
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Skidding on a Curve: Unbanked Curves – Uniform Circular Motion Example Problem

A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s Will the car follow the curve, or will it skid? Assume: (a) the pavement is dry and the coefficient of static friction is \mu _s=0.60; (b) the pavement is icy and \mu _s=0.25.

Solution:

The forces on the car are gravity mg downward, the normal force FN exerted upward by the road, and a horizontal friction force due to the road. They are shown in the free-body diagram of the car below. The car will follow the curve if the maximum static friction force is greater than the mass times the centripetal acceleration.

Part A

In the vertical direction (y) there is no acceleration. Newton’s second law tells us that the normal force on the car is equal to the weight mg since the road is flat:

\begin{align*}
\sum_{}^{}F_y & =m\ a_c\\
\\
F_N\ -\ mg & =0\\
\\
F_N&=mg\\
\\
F_N &=\left( 1000\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)\\
\\
F_N &=9810\  \text{N}
\end{align*}

In the horizontal direction the only force is friction, and we must compare it to the force needed to produce the centripetal acceleration to see if it is sufficient. The net horizontal force required to keep the car moving in a circle around the curve is

\begin{align*}
\sum_{}^{}F_c & =m\ a_c\\
\\
 & =m\cdot \frac{v^2}{r}\\
\\
& =\left( 1000\ \text{kg} \right)\cdot \frac{\left( 15\ \text{m/s} \right)^2}{50\ \text{m}}\\
\\
&=4500\ \text{N}
\end{align*}

Now we compute the maximum total static friction force (the sum of the friction forces acting on each of the four tires) to see if it can be large enough to provide a safe centripetal acceleration. For (a), \mu _s=0.60, and the maximum friction force attainable is

\begin{align*}
\sum_{}^{}F_{fr_{max}}& =\mu _s \ F_N\\
\\
&=\left( 0.60 \right)\left( 9810\ \text{N} \right)\\
\\
&=5886\ \text{N}
\end{align*}

Since a force of only 4500 N is needed, and that is, in fact, how much will be exerted by the road as a static friction force, the car can follow the curve.

Part B

The maximum static friction force possible is

\begin{align*}
\sum_{}^{}F_{fr_{max}}& =\mu _s \ F_N\\
\\
&=\left( 0.25 \right)\left( 9810\ \text{N} \right)\\
\\
&=2452.5\ \text{N}
\end{align*}

The car will skid because the ground cannot exert sufficient force (4500 N is needed) to keep it moving in a curve of radius 50 m at a speed of 54 km/h.