College Physics 2.64 – Consistency on the position and velocity diagrams


(a) Take the slope of the curve in Figure 2.64 to find the jogger’s velocity at t = 2.5 s.

(b) Repeat at 7.5 s. These values must be consistent with the graph in Figure 2.65.

Figure 2.64
Figure 2.65

Solution:

The velocity at time 2.5 seconds is 5 m/s.

The velocity at time 7.5 seconds is -4 m/s.


College Physics 2.63 – Constructing a displacement graph


Construct the displacement graph for the subway shuttle train as shown in Figure 2.18(a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.

2.63


Solution:

The position vs time graph is shown in the figure below. 

2.63b

College Physics 2.62 – Acceleration as the slope of the velocity vs time diagram


By taking the slope of the curve in Figure 2.63, verify that the acceleration is 3.2 m/s²  at t = 10 s.

2.62
Figure 2.63

Solution:

Since the graph is a straight line, we can use the two points before and after the specified time to determine the slope of the line. The slope of the velocity-time graph is the acceleration.

The two points are \left(0,\:165\right)\:and\:\left(20,\:228\right)

The velocityis computed as

a=m=\frac{\Delta y}{\Delta x}

a=\frac{y_2-y_2}{x_2-x_1}

a=\frac{228\:m/s-165\:m/s}{20\:s-0\:s}

a=3.2\:m/s^2

Therefore, the acceleration is verified to be 3.2 m/s² at t=10 s.


College Physics 2.61 – Velocity as slope of position vs time


Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t = 30.0 s is 0.238 m/s. Assume all values are known to 3 significant figures.

2.60
Figure 2.62

Solution:

Since the graph is a straight line, we can use any two points to determine the slope of the line. The slope of the position-time graph is the velocity.

The two points are \left(20,\:6.95\right)\:and\:\left(40,\:11.7\right)

The velocityis computed as

v=m=\frac{\Delta y}{\Delta x}

v=\frac{y_2-y_2}{x_2-x_1}

v=\frac{11.7\:km-6.95\:km}{40\:s-20\:s}

v=0.238\:km/s

Therefore, the velocity is 0.238 km/s.


College Physics 2.60 – Slope of the position vs time diagram


Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t = 10.0 s is 0.208 km/s. Assume all values are known to 3 significant figures.

2.60
Figure 2.62

Solution:

Since the graph is a straight line, we can use any two points to determine the slope of the line. The slope of the position-time graph is the velocity.

The two points are \left(0,\:2.80\right)\:and\:\left(20,\:6.95\right)

The velocityis computed as

v=m=\frac{\Delta y}{\Delta x}

v=\frac{y_2-y_2}{x_2-x_1}

v=\frac{6.95\:km-2.80\:km}{20\:s-0\:s}

v=0.208\:km/s

Therefore, the velocity is 0.208 km/s.


College Physics 2.59 – Analysis of motion diagrams


(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

Based from the graph shown in figure 2.60, we can choose two points, one before and one after t=20 s. We have the points \left(15,\:988\right)\:and\:\left(25,\:2138\right).

The slope is computed using the formula:

m=\frac{\Delta y}{\Delta x}

m=\frac{2138\:m-988\:m}{25\:s-15\:s}

m=115\:m/s

Therefore, we have verified that the velocity of the jet car is 115 m/s at t=20 seconds. 

Part B

Since the graph is a straight line, we can use any two points to determine the slope of the line. The slope of the velocity-time graph is the acceleration.

The two points are \left(10,\:65\right)\:and\:\left(25,\:140\right)

The acceleration is computed as

a=m=\frac{\Delta y}{\Delta x}

a=\frac{y_2-y_2}{x_2-x_1}

a=\frac{140\:m/s-65\:m/s}{25\:s-10\:s}

a=5\:m/s^2

Therefore, the acceleration is verified to be 5.0 m/s².


College Physics 2.58 – A tennis ball dropped onto a hard floor


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 (0.0035 s).

(d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?


Solution:

Part A

This problem is the same as Problem 2.56. The velocity of the ball when it strikes the ground is

v=\sqrt{\left(v_o\right)^2+2a\Delta y}

v=\sqrt{\left(0\:m/s\right)^2+2\left(-9.80\:m/s^2\right)\left(-1.50\:m\right)}

v=5.42\:m/s\:\left(downward\right)

Part B

The velocity of the ball just after it leaves the floor on its way back up

v_o=\sqrt{v^2-2a\Delta y}

v_o=\sqrt{\left(0\:m/s\right)^2-2\left(-9.80\:m/s^2\right)\left(1.10\:m\right)}

v_o=4.64\:m/s upward

Part C

The acceleration is

a=\frac{v-v_o}{t}

a=\frac{4.643\:m/s-\left(-5.422\:m/s\right)}{0.0035\:s}

a=2880\:m/s^2

Part D

\Delta y=\frac{\left(v_o\right)^2-v^2}{2a}

\Delta y=\frac{\left(-5.422\:m/s\right)^2-\left(0\:m/s\right)^2}{2\left(2880\:m/s^2\right)}

\Delta y=0.0051\:m


College Physics 2.57 – A coin dropped from a hot-air balloon


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

For the coin, we are given the following:  v_o=+10\:m/s;\:a=-9.80\:m/s^2;\:v=0\:m/s\:\left(at\:top\:of\:ascent\right)

The maximum height reached can be solved using the formula

v^2=\left(v_o\right)^2+2a\Delta y

Solve for Δy in terms of the other variables

\Delta y=\frac{v^2-\left(v_o\right)^2}{2a}

Now, substitute the given values

\Delta y=\frac{\left(0\:m/s\right)^2-\left(10\:m/s\right)^2}{2\left(-9.80\:m/s\right)^2}

\Delta y=5.10\:m

From the moment the coin is dropped to the top of the ascent, the change in position of the coin is +5.10 m. Therefore, the maximum height is 

y_{max}=300\:m+5.10\:m

y_{max}=305.10\:m

Part B

The position of the coin after 4 seconds can be solved using the formula

\Delta y=v_ot+\frac{1}{2}at^2

Substitute the given values

y=\left(10\:m/s\right)\left(4\:s\right)+\frac{1}{2}\left(-9.80\:m/s^2\right)\left(4\:s\right)^2

y=-38.4\:m

The change in position of the coin is -38.4 m. Therefore, the poisiton of the coin with respect to the ground is

y_{_{\left(4\:s\right)}}=300\:m-38.4\:m

y_{_{\left(4\:s\right)}}=261.6\:m

The velocity after 4 seconds is given by the formula

v=v_o+at

Substitute the given values

v=10\:m/s+\left(-9.8\:m/s^2\right)\left(4.00\:s\right)

v=-29.2\:m/s

The velocity is 29.2 m/s, downward. 

Part C

The time it takes for the coin to hit the ground can be solved using the formula

\Delta y=v_ot+\frac{1}{2}at^2

Substitute the given values

300\:m=\left(10.0\:m/s\right)t+\frac{1}{2}\left(-9.80\:m/s^2\right)t^2

Simplify and rearrange the equation to form a quadratic equation

-4.90t^2+10t-300=0

Now, we can solve for time, t, using the quadratic formula

t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

t=\frac{-\left(10\right)\pm \sqrt{\left(10\right)^2-4\left(-4.9\right)\left(300\right)}}{2\left(-4.9\right)}

t=8.91\:s

The time of flight is 8.91 seconds.


College Physics 2.56 – A steel ball drops and rebounds onto a hard floor


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.

(d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?


Solution:

Part A

We are given an initial velocity of 0. We are required to solve for the final velocity. We shall use the formula

v^2=\left(v_o\right)^2+2a\Delta y

Then, we substitute the given values

v^2=\left(0\:m/s\right)^2+2\left(-9.80\:m/s^2\right)\left(0\:m-1.50\:m\right)

v=5.42\:m/s

The velocity is 5.42 m/s downward. 

Part B

Consider the floor to be the initial position and the top of the trajectory to be the final position. We shall use the formula

v^2=\left(v_o\right)^2+2a\Delta y

The initial velocity is the unknown, so we solve the equation in terms of the initial velocity. 

v_o=\sqrt{v^2-2a\Delta y}

Substitute the given values

v_o=\sqrt{\left(0\:m/s\right)^2-2\left(-9.80\:m/s^2\right)\left(1.45\:m\right)}

v_o=5.33\:m/s

The initial velocity is 5.33 m/s upward. 

Part C

The acceleration is calculated using the formula

v=v_o+at

Solve for the equation in terms of acceleration

a=\frac{v-v_o}{t}

Substitute the given values

a=\frac{5.331\:m/s-\left(-5.422\:m/s\right)}{8.00\times 10^{-5}\:s}

a=1.34\times 10^5\:m/s^2

Part D

The period of compression occurs when the ball goes from -5.42 m/s to 5.33 m/s. The acceleration during this time is found in part C. The compression is solve using the formula 

v^2=\left(v_o\right)^2+2a\Delta y

Solve for Δy in terms of the other variables

\Delta y=\frac{\left(v^2\right)-\left(v_o\right)^2}{2a}

Then we substitute the given values

\Delta y=\frac{\left(5.42\:m/s\right)^2-\left(0\:m/s\right)^2}{2\left(1.34\times 10^5\:m/s^2\right)}

\Delta y=-1.09\times 10^{-4}\:m


College Physics 2.55 – A rock dropped into a dark well


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

So, we just need to solve for the distance y using the equation

y=v_ot+\frac{1}{2}at^2

Substitute the given values

y=0+\frac{1}{2}\left(-9.80\:m/s^2\right)\left(2.0\:s\right)^2=-19.6\:m

Neglecting the speed of sound, the distance to the water is 19.6 m below the ground.

Part B

Let h be the depth of the well, y_o=0\:m,\:y=-h at the bottom of the well. Let v_2 be the speed of sound and t_2 be the time for the sound to travel from the bottom to the top of the well. Let t_1 be the time for the rock to reach the bottom.

t_1+t_2=2.0\:s                                         Equation1

-h=\frac{1}{2}a\left(t_1\right)^2                          Equation2

h=v_2t_2=v_2\left(2.0-t_1\right)                         Equation 3

Adding equation 2 and 3

0=\frac{1}{2}a\left(t_1\right)^2+v_2\left(2.0-t_1\right)

Substitute the given values

0=0.5\left(-9.8\right)\left(t_1\right)^2+\left(332\right)\left(2-t_1\right)

0=-4.9\left(t_1\right)^2+664-332t_1

Rearranging, we have

-4.9\left(t_1\right)^2-332t_1\:+664=0

Solve the equation using the quadratic formula

t_1=\frac{-\left(-332\right)\pm \sqrt{\left(-332\right)^2-4\left(-4.90\right)\left(664\right)}}{2\left(-4.9\right)}=1.944\:s

We have solved for t_1. Now, we can substitute this value to equation 2 to solve for h

h=-\frac{1}{2}\left(-9.80\right)\left(1.944\right)^2=18.5\:m

With the speed of sound taken into account, the distance to the water is about 18.5 meters.