Tag Archives: Kinematics

College Physics by Openstax Chapter 2 Problem 65

A graph of v(t) is shown for a world-class track sprinter in a 100-m race. (See Figure 2.67). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at t=5 s? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?

Solution:

Part A

To find the average velocity over the straight line graph of the velocity vs time shown, we just need to locate the midpoint of the line. In this case, the average speed for the first 4 seconds is

v_{\text{ave}}=6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

Looking at the graph, the velocity at exactly 5 seconds is 12 m/s.

Part C

If we are given the velocity-time graph, we can solve for the acceleration by solving for the slope of the line.

Consider the line from time zero to time, t=4 seconds. The slope, or acceleration, is

a=\text{slope}=\frac{12\:\text{m/s}-0\:\text{m/s}}{4\:\text{s}}=3\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part D

For the first 4 seconds, the distance traveled is equal to the area under the curve.

\text{distance}=\frac{1}{2}\left(4\:\sec \right)\left(12\:\text{m/s}\right)=24\:\text{m}

So, the sprinter traveled a total of 24 meters in the first 4 seconds. He still needs to travel a distance of 76 meters to cover the total racing distance. At the constant rate of 12 m/s, he can run the remaining distance by

\text{t}=\frac{\text{distance}}{\text{velocity}}=\frac{76\:\text{m}}{12\:\text{m/s}}=6.3\:\sec

Therefore, the total time of the sprint is

\text{t}_{\text{total}}=4\:\sec +6.3\:\sec =10.3\:\sec \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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Grantham PHY220 Week 2 Assignment Problem 8

If a car is traveling at 50 m/s and then stops over 300 meters (while sliding), what is the coefficient of kinetic friction between the tires of the car and the road?

SOLUTION:

Draw the free-body diagram of the car

Consider the vertical direction

$\sum F_y=ma_y$

$F_N-mg=0$

$F_N=mg$

Consider the motion in the horizontal direction

Solve for the acceleration of the car.

$v^2=\left(v_0\right)^2+2a_x\Delta x$

$a_x=\frac{v^2-\left(v_0\right)^2}{2\Delta x}=\frac{0-50^2}{2\left(300\right)}=-4.17\:m/s^2$

Solve for the coefficient of kinetic friction

$\sum F_x=ma_x$

$-F_{fr}=ma_x$

$-\mu _kF_N=ma_x$

$\mu _k=\frac{ma_x}{-F_N}=\frac{m\:\left(-4.17\right)}{-m\left(9.80\right)}=\frac{4.17}{9.80}$

$\mu _k=0.43$

Grantham PHY220 Week 2 Problem 3

A bullet is fired from a gun at a shooting range. The bullet hits the ground after 0.32 seconds. How far did it travel horizontally and vertically in this time if it was fired at a velocity of 1100 m/s?

Solution:

\Delta x=V_{0_x}t=\left(1100\:m/s\right)\left(0.32\:s\right)=352\:m

College Physics by Openstax Chapter 2 Problem 49

You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass before the ball passes the tree branch on the way back down?

Solution:

The known values are $a=-9.80\:\text{m/s}^2$ ; $v_o=15.0\:\text{m/s}$ ; $y=7.00\:\text{m}$

The applicable formula is.

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

Using this formula, we can solve it in terms of time, $t$ .

t=\frac{-v_0\pm \sqrt{v_0^2+2ay}}{a}

Substituting the known values, we have

\begin{align*} t & =\frac{-v_0\pm \sqrt{v_0^2+2ay}}{a} \\ t & =\frac{-15.0\:\text{m/s}\pm \sqrt{\left(15.0\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(7.00\:\text{m}\right)}}{-9.80\:\text{m/s}^2} \\ t&=\frac{-15.0\:\text{m/s}\pm 9.37\:\text{m/s}}{-9.80\:\text{m/s}^2} \end{align*}

We have two values for time, $t$ . These two values represent the times when the ball passes the tree branch.

t_1=\frac{-15.0\:m/s+9.37\:m/s}{-9.80\:m/s^2}=0.57\:sec \\ t_2=\frac{-15.0\:m/s-9.37\:m/s}{-9.80\:m/s^2}=2.49\:sec

Therefore, the total time between passing the branch is the difference between 2.49 seconds and 0.57 seconds.

t_2-t_1=2.49 \ \text{s} - 0.57 \ \text{s}=1.92 \ \text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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College Physics by Openstax Chapter 2 Problem 48

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long does he have to get out of the way if the shot was released at a height of 2.20 m, and he is 1.80 m tall?

Solution:

The known values are: $y_0=2.20\:\text{m}$ ; $y=1.80\:\text{m}$ ; $v_0=11.0\:\text{m/s}$ ; and $a=-9.80\:\text{m/s}^2$

We are going to use the formula

\Delta y=v_0t+\frac{1}{2}at^2

Substituting the given values:

\begin{align*} \Delta y & =v_0t+\frac{1}{2}at^2 \\ 1.80\:\text{m}-2.20\:\text{m} & =\left(11.0\:\text{m/s}\right)t+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)t^2 \\ -0.40\:\text{m} & =\left(11.0\:\text{m/s}\right)t-\left(4.90\:\text{m/s}^2\right)t^2 \\ 4.90t^2-11t-0.40 & =0 \end{align*}

Using the quadratic formula solve for $t$ , we have

\begin{align*} t & =\frac{-\left(-11\right)\pm \sqrt{\left(-11\right)^2-4\left(4.90\right)\left(-0.40\right)}}{2\left(4.90\right)} \\ \end{align*}

t=2.28\:\text{sec}\:\:\:\:\:\text{or}\:\:\:\:\:\:t=-0.04 \ \text{sec}

We can discard the negative time, so

t=2.28\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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College Physics by Openstax Chapter 2 Problem 47

(a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s.

(b) How long would it take to reach the ground if it is thrown straight down with the same speed?

Solution:

Part A

Refer to the figure below.

The known values are: $t=2.35\:\text{s}$ ; $y=0\:\text{m}$ ; $v_0=+8.00\:\text{m/s}$ ; and $a=-9.8\:\text{m/s}^2$

Based on the given values, the formula that we shall use is

y=y_0+v_0t+\frac{1}{2}at^2

Substituting the values, we have

\begin{align*} y & =y_0+v_0t+\frac{1}{2}at^2 \\ 0\: & =y_0+\left(8.00\:\text{m/s}\right)\left(2.35\:\text{s}\right)+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)\left(2.35\:\text{s}\right)^2 \\ y_0 & =8.26\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}

Therefore, the cliff is 8.26 meters high.

Part B

Refer to the figure below

The knowns now are: $y=0\:\text{m}$ ; $y_0=8.26\:\text{m}$ ; $v_0=-8.00\:\text{m/s}$ ; and $a=-9.80\:\text{m/s}^2$

Based on the given values, we can use the formula

y=y_0+v_0t+\frac{1}{2}at^2

Substituting the values, we have

\begin{align*} y & =y_0+v_0t+\frac{1}{2}at^2 \\ 0\:\text{m} & =8.26\:\text{m}+\left(-8.00\:\text{m/s}\right)t+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)t^2 \\ 4.9 t^2+8t-8.26 & =0 \\ \end{align*}

Using the quadratic formula to solve for the value of $t$ , we have

\begin{align*} t &=\frac{-8\pm \sqrt{\left(8\right)^2-4\left(4.9\right)\left(-8.26\right)}}{2\left(4.9\right)} \\ t &=0.717\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics by Openstax Chapter 2 Problem 46

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool.

(a) How long are her feet in the air?

(b) What is her highest point above the board?

(c) What is her velocity when her feet hit the water?

Solution:

The known values are: $y_0=1.80\:\text{m}$ , $y=0\:\text{m}$ , $a=-9.80\:\text{m/s}^2$ , $v_0=4.00\:\text{m/s}$ .

Part A

Based from the knowns, the formula most applicable to solve for the time is $\Delta y=v_0t+\frac{1}{2}at^2$ . If we rearrange the formula by solving for $t$ , and substitute the given values, we have

\begin{align*} t & =\frac{-v_0\pm \sqrt{v_0^2-2a\Delta y}}{a} \\ t & =\frac{-4.00\:\text{m/s}\pm \sqrt{\left(4.00\:\text{m/s}\right)^2-2\left(-9.80\:\text{m/s}^2\right)\left(1.80\:\text{m}\right)}}{-9.80\:\text{m/s}^2} \\ t & =1.14\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We have the formula

\begin{align*} \Delta y & =\frac{v^2-v_0^2}{2a} \\ \Delta y & =\frac{\left(0\:\text{m/s}\right)^2-\left(4.00\:\text{m/s}\right)^2}{2\left(-9.80\:\text{m/s}^2\right)} \\ \Delta y & =0.816\: \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The formula to be used is

v^2=v_0^2+2a\Delta y

Substituting the given values

\begin{align*} v^2 & =v_0^2+2a\Delta y \\ v & =\pm \sqrt{v_0^2+2a\Delta y} \\ v & =\pm \sqrt{\left(4.00\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(-1.80\:\text{m}\right)} \\ v & =\pm \sqrt{51.28\:\text{m}^2/\text{s}^2} \\ v &=\pm 7.16\:\text{m/s} \end{align*}

Since the diver must be moving in the negative direction,

v=-7.16\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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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 43

A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?

Solution:

It is our assumption that the player attempts to get the ball at the top where the velocity is zero.

The given are the following: $v_{fy}=0 \ \text{m/s}$ ; $\Delta y=1.25 \ \text{m}$ ; and $a=-9.80 \ \text{m/s}^2$ .

We are required to solve for the initial velocity $v_{0y}$ of the player. We are going to use the formula

\left(v_{fy}\right)^2=\left(v_{oy}\right)^2+2a\Delta y

Solving for $v_{oy}$ in terms of the other variables:

v_{oy}=\sqrt{\left(v_{fy}\right)^2-2a\Delta y}

Substituting the given values:

\begin{align*} v_{oy} & =\sqrt{\left(v_{fy}\right)^2-2a\Delta y} \\ v_{oy} & = \sqrt{\left(0\:\text{m/s}\right)^2-2\left(-9.80\:\text{m/s}^2\right)\left(1.25\:\text{m}\right)} \\ v_{oy} & =4.95 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Tag Archives: Kinematics

College Physics by Openstax Chapter 2 Problem 65

A graph of v(t) is shown for a world-class track sprinter in a 100-m race. (See Figure 2.67). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at t=5 s? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?

Grantham PHY220 Week 2 Assignment Problem 8

If a car is traveling at 50 m/s and then stops over 300 meters (while sliding), what is the coefficient of kinetic friction between the tires of the car and the road?

Grantham PHY220 Week 2 Problem 3

A bullet is fired from a gun at a shooting range. The bullet hits the ground after 0.32 seconds. How far did it travel horizontally and vertically in this time if it was fired at a velocity of 1100 m/s?

College Physics by Openstax Chapter 2 Problem 49

You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass before the ball passes the tree branch on the way back down?

College Physics by Openstax Chapter 2 Problem 48

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long does he have to get out of the way if the shot was released at a height of 2.20 m, and he is 1.80 m tall?

College Physics by Openstax Chapter 2 Problem 47

(a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s.

(b) How long would it take to reach the ground if it is thrown straight down with the same speed?

Part B

College Physics by Openstax Chapter 2 Problem 46

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool.

(a) How long are her feet in the air?

(b) What is her highest point above the board?

(c) What is her velocity when her feet hit the water?

College Physics by Openstax Chapter 2 Problem 45

College Physics by Openstax Chapter 2 Problem 44

College Physics by Openstax Chapter 2 Problem 43

A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?