A Flower Pot Falling Past a Window| University Physics


As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time \displaystyle  t, and the vertical length of your window is \displaystyle L_w. Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity \displaystyle g.

Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward).

PART A. From what height \displaystyle h above the bottom of your window was the flower pot dropped?

PART B. If the bottom of your window is a height \displaystyle h_b above the ground, what is the velocity \displaystyle v_{ground} of the pot as it hits the ground? You may introduce the new variable \displaystyle v_b, the speed at the bottom of the window, defined by

\displaystyle v_b=\frac{gt}{2}+\frac{L_w}{t}


Solution:

Part A

The initial velocity of the pot is zero. Find the velocity \displaystyle v_b of the pot at the bottom of the window. Then using the kinematic equation that relates initial and final velocities, acceleration, and distance traveled, you can solve for the distance \displaystyle h.

The average velocity of the flower pot as it passes by the window is \displaystyle v_{avg}=\frac{L_w}{t}.

As the pot falls past your window, there will be some instant when the pot’s velocity equals the average velocity \displaystyle v_{avg}. Recall that, under constant acceleration, velocity changes linearly with time. This means that the average velocity during a time interval will occur at the middle of that time interval. Meaning, the average velocity happened at time, \displaystyle \frac{t}{2}.

Considering the motion at the middle and at the bottom of the window.

\displaystyle g=\frac{v_b-v_{avg}}{\frac{t}{2}}

\displaystyle v_b=\frac{gt}{2}+v_{ave}

\displaystyle v_b=\frac{gt}{2}+\frac{L_w}{t}

So, we now know that velocity at the bottom of the window. Consider the motion from the top (point of dropped) to the bottom of the window

\displaystyle \left(v_b\right)^2-\left(v_0\right)^2=2g\left(h-0\right)

\displaystyle \left(v_b\right)^2=2gh

\displaystyle h=\frac{\left(v_b\right)^2}{2g}

\displaystyle h=\frac{\left(\frac{gt}{2}+\frac{L_w}{t}\right)^2}{2g}=\frac{\left(\frac{gt^2+2L_w}{2t}\right)^2}{2g}=\frac{\left(gt^2+2L_w\right)^2}{8gt^2}

Part B

\displaystyle \left(v_{ground}\right)^2-\left(v_b\right)^2=2gh_b

\displaystyle \left(v_{ground}\right)^2=2gh_b+\left(v_b\right)^2

\displaystyle\left(v_{ground}\right)=\sqrt{2gh_b+\left(v_b\right)^2}


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To demonstrate the tremendous acceleration of a top fuel Drag Racer| University Physics


To demonstrate the tremendous acceleration of a top fuel drag racer, you attempt to run your car into the back of a dragster that is “burning out” at the red light before the start of a race. (Burning out means spinning the tires at high speed to heat the tread and make the rubber sticky.)

You drive at a constant speed of \displaystyle v_0 toward the stopped dragster, not slowing down in the face of the imminent collision. The dragster driver sees you coming but waits until the last instant to put down the hammer, accelerating from the starting line at constant acceleration, \displaystyle a. Let the time at which the dragster starts to accelerate be \displaystyle t=0.

The figure shows a car moving to the right with speed v 0 toward a stopped dragster, which stands in front of the lights facing to the right.

PART A. What is \displaystyle t_{max}, the longest time after the dragster begins to accelerate that you can possibly run into the back of the dragster if you continue at your initial velocity?


Answer:

\displaystyle t_{max}=\frac{v_0}{a}


PART B. Assuming that the dragster has started at the last instant possible (so your front bumper almost hits the rear of the dragster at \displaystyle t=t_{max}), find your distance from the dragster when he started. If you calculate positions on the way to this solution, choose coordinates so that the position of the drag car is 0 at t=0. Remember that you are solving for a distance (which is a magnitude, and can never be negative), not a position (which can be negative).


Answer:

\displaystyle D_{start}=\frac{\left(v_0\right)^2}{2a}

\displaystyle D_{start}=D_{car}-x_d\left(t_{max}\right)

\displaystyle D_{start}=v_0t_{max}-\frac{1}{2}a\left(t_{max}\right)^2

\displaystyle D_{start}=v_0\left(\frac{v_0}{a}\right)-\frac{1}{2}a\left(\frac{v_0}{a}\right)^2

\displaystyle D_{start}=\frac{\left(v_0\right)^2}{a}-\frac{\left(v_0\right)^2}{2a}

\displaystyle D_{start}=\frac{\left(v_0\right)^2}{2a}


PART C. Find numerical values for \displaystyle t_{max} and \displaystyle D_{start} in seconds and meters for the (reasonable) values \displaystyle v_0=60\:mph (26.8 m/s) and \displaystyle a=50\:m/s^2


Answer:

\displaystyle t_{max}=\frac{v_0}{a}=\frac{26.8\:m/s}{50\:m/s^2}=0.54\:s

\displaystyle D_{start}=\frac{\left(v_0\right)^2}{2a}=\frac{\left(26.8\:m/s\right)^2}{2\left(50\:m/s^2\right)}=7.2\:m

The blue curve shows how the car, initially at \displaystyle x_0, continues at constant velocity (blue) and just barely touches the accelerating drag car (red) at \displaystyle t_{max}.


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Running and Walking Problem| University Physics


Tim and Rick both can run at speed \displaystyle v_r and walk at speed \displaystyle v_w , with \displaystyle v_r>v_w. They set off together on a journey of distance D. Rick walks half of the distance and runs the other half. Tim walks half of the time and runs the other half.

PART A. How long does it take Rick to cover the distance D?


Answer:

Find the time that it takes Rick to walk the first half of the distance, that is, to travel a distance D/2 at speed \displaystyle v_w.

\displaystyle t_{w,R}=\frac{D}{2v_w}

Now find the time Rick spends running.

\displaystyle t_{r,R}=\frac{D}{2v_r}

Now just add the two times up and you’re done.

\displaystyle t_R=\frac{D}{2v_w}+\frac{D}{2v_r}=\frac{D}{2v_wv_r}\left(v_w+v_r\right)


PART B. Find Rick’s average speed for covering the distance D.


Answer:

You were given the total distance and have calculated the total time. Recall that average speed is equal to the total distance traveled divided by the amount of time it took to travel this distance.

\displaystyle v_{ave,\:R}=\frac{2v_rv_w}{v_w+v_r}


PART C. How long does it take Tim to cover the distance?


Answer:

Tim walks at speed \displaystyle v_w half the time and runs at speed \displaystyle v_r for the other half.

\displaystyle v_{ave,\:T}=\frac{v_w+v_r}{2}

The time is just the distance divided by the average speed.

\displaystyle t_T=\frac{D}{\frac{v_w+v_r}{2}}=\frac{2D}{v_r+v_w}


PART D. Who covers the distance D more quickly?


Answer: Tim

Imagine that both Rick and Tim do all of their walking before they start to run. Rick will start running when he has covered half of the total distance. When Tim reaches half of the total distance, will he already have started running?


PART E. In terms of given quantities, by what amount of time, Δt, does Tim beat Rick?


Answer:

\displaystyle \Delta t=\frac{D\left(v_w-v_r\right)^2}{2v_rv_w\left(v_r-v_w\right)}

This is just simple subtraction between the two computed times.


PART F. In the special case that vr=vw, what would be Tim’s margin of victory Δt(vr=vw)?


ANSWER: 0

If vr=vw, is the any difference between what Tim and Rick do?


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Half the Distance and Half the Time Problem| University Physics


Julie drives 100 mi to Grandmother’s house. On the way to Grandmother’s, Julie drives half the distance at 35.0 mph and half the distance at 65.0 mph. On her return trip, she drives half the time at 35.0 mph and half the time at 65.0 mph.

PART A. What is Julie’s average speed on the way to Grandmother’s house?


ANSWER: 45.5 mph

Julie drove 50 miles at a speed of 35 mph, and drove another 50 miles for 65 mph. So, for the first 50 miles, she drove for

\displaystyle time=\frac{distance}{speed}=\frac{50\:mi}{35\:mph}=\frac{10}{7}\:hours

and for the next 50 miles, she drove for

\displaystyle time=\frac{distance}{speed}=\frac{50\:mi}{65\:mph}=\frac{10}{13}\:hours

Therefore, her average speed was

\displaystyle average\:speed=\frac{total\:distance}{total\:time}

\displaystyle average\:speed=\frac{100\:miles}{\frac{10}{7}+\frac{10}{13}\:hours}=45.5\:mph


PART B. What is her average speed on the return trip?


ANSWER: 50.0 m/s

Since the time she used driving at 35 mph is the same amount of time she used driving at 65 mph, the average speed is just the average of the two speeds given.


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Position, Velocity, and Acceleration values from Position Function| University Physics


A particle moving along the x-axis has its position described by the function \displaystyle x=\left(2.00t^3-5.00t+3.00\right)m, where t is in s. At t= 3.00, what are the particle’s (a) position, (b) velocity, and (c) acceleration?


PART A.

ANSWER: 42.0 m

Evaluate the position at time t= 3.00 s.

\displaystyle x=2.00\left(3\right)^3-5.00\left(3\right)+3=42.0\:m


PART B.

ANSWER: 49.0 m/s

Determine the velocity function v(t) from the position function x(t) by differentiation.

\displaystyle v\left(t\right)=6t^2-5

\displaystyle v\left(t\right)=6\left(3\right)^2-5=49.0\:m/s


PART C.

ANSWER: 36.0 m/s²

Determine the acceleration function a(t) from the the velocity function by differentiation

\displaystyle a\left(t\right)=12t

\displaystyle a\left(t\right)=12t=12\left(3\right)=36\:m/s^2


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Analyzing Position versus Time Graphs: Conceptual Question| University Physics

Problem:

Two cars travel on the parallel lanes of a two-lane road. The cars’ motions are represented by the position versus time graph shown in the figure. Answer the questions using the times from the graph indicated by letters.

Two graphs for positions of two cars, labeled car 1 and car 2, as functions of time are shown on the same set of axes. Time is measured on the positive x-axis and the position is measured on the y-axis. A grid is shown on the plane. The position of car 1 increases linearly from the point with coordinates (0,-3) to the point with approximate coordinates of (11.5, 4). Graph for car 1 crosses the x-axis at point with coordinates (5, 0) at time B. The position of car 2 smoothly increases from the point with approximate coordinates of (0, -0.5), crossing the x-axis at (1, 0) at time A, to the maximum at point with coordinates (6, 4) at time C, when the slope of the graph equals zero. Then the position smoothly decreases to a point with approximate coordinates of (12, -1), crossing the graph for car 1 at (9, 2.5) at time D, and crossing an x-axis again at (11, 0), at time E.

PART A. At which of the times do the two cars pass each other?

ANSWER: D

Two objects can pass each other only if they have the same position at the same time.

PART B. Are the two cars traveling in the same direction when they pass each other?

ANSWER: No

PART C. At which of the lettered times, if any, does car #1 momentarily stop?

ANSWER: None

The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed.

PART D. At which of the lettered times, if any, does car #2 momentarily stop?

ANSWER: C

PART E. At which of the lettered times are the cars moving with nearly identical velocity?

ANSWER: A

Position versus Time Graphs Conceptual Question| University Physics

Problem:

The motions described in each of the questions take place at an intersection on a two-lane road with a stop sign in each direction. For each motion, select the correct position versus time graph. For all of the motions, the stop sign is at the position x=0, and east is the positive x-direction.

The figure shows six graphs, labeled A to Feach of which shows the position as a function of time. Time is measured on the positive x-axis and the position is measured on the y-axis. On graph A the position decreases linearly from the point, which lies on the positive y-axis, intersecting the positive x-axis. On graph B the position increases linearly from the origin. On graph C the position smoothly increases from the origin, forming a concave curve and reaching a constant value. On graph D the position increases linearly from the point, which lies on the negative y-axisintersecting the positive x-axis. On graph E the position smoothly increases from the origin, forming a convex curve. On graph F the position smoothly decreases from the point, which lies on the positive y-axis, to zeroforming a convex curve.

PART A. A driver ignores the stop sign and continues driving east at constant speed.

ANSWER: D

PART B. A driver ignores the stop sign and continues driving west at constant speed.

ANSWER: A

PART C. A driver, traveling west, slows and stops at the stop sign.

ANSWER: F

PART D. A driver, after stopping at the stop sign, travels east with a positive acceleration.

ANSWER: E

Arizona State University| PHY 121: Univ Physics I: Mechanics|Homework 1-2| Velocity from Graphs of Position versus Time

Problem:

An object moves along the x-axis during four separate trials. Graphs of position versus time for each trial (with the same scales on each axis) are shown in the figure.

The figure shows four graph, labeled trial A, trial B, trial C and trial Deach of which shows the position as a function of time. Time is measured on the positive x-axis and the position is measured on the y-axis. On graph trial A the position increases linearly from the origin. On graph trial B the position smoothly increases from the point, which lies on the negative y-axis, forming a convex curve and intersecting the positive x-axis. On graph trial C the position remains constant at some positive value. On graph trial D the position decreases linearly from the point, which lies on the positive y-axis intersecting the positive x-axis.

PART A. During which trial or trials is the object’s velocity not constant?

ANSWER: Trial B

The graph of the motion during Trial B has a changing slope and therefore is not constant. The other trials all have graphs with constant slope and thus correspond to motion with constant velocity.

PART B. During which trial or trials is the magnitude of the average velocity the largest?

ANSWER: Trials B and D

While Trial B and Trial D do not have the same average velocity, the only difference is the direction! The magnitudes are the same. Neither one is “larger” than the other, and it is only because of how we chose our axes that Trial B has a positive average velocity while Trial D has a negative average velocity. In Trial C the object does not move, so it has an average velocity of zero. During Trial A the object has a positive average velocity but its magnitude is less than that in Trial B and Trial D.

Arizona State University| PHY 121: Univ Physics I: Mechanics|Homework 1-2| Average Velocity from a Position vs. Time Graph

Learning Goal: To learn to read a graph of position versus time and to calculate average velocity.

In this problem, you will determine the average velocity of a moving object from the graph of its position x(t) as a function of time t. A traveling object might move at different speeds and in different directions during an interval of time, but if we ask at what constant velocity the object would have to travel to achieve the same displacement over the given time interval, that is what we call the object’s average velocity. We will use the notation vave[t1,t2] to indicate average velocity over the time interval from t1 to t2. For instance, vave[1,3] is the average velocity over the time interval from t=1 to t=3.

Problem:

PART A. Consulting the graph shown in the figure, find the object’s average velocity over the time interval from 0 to 1 second.

The plot shows the position as a function of time. Time is measured from 0 to 6 seconds on the x-axis. The position is measured from 0 to 60 meters on the y-axis. The position remains constant at a value of 20 meters from 0 to 1 second. Then, it increases linearly to 3 seconds and 60 meters. Then, it remains constant at a value of 60 meters to 4.5 seconds. Then, the position decreases linearly to 6 seconds and 20 meters.

ANSWER: 0 m/s

Average velocity is defined as the constant velocity at which an object would have to travel to achieve a given displacement (difference between final and initial positions, which can be negative) over a given time interval, from the initial time ti to the final time tf. The average velocity is therefore equal to the displacement divided by the given time interval. In symbolic form, average velocity is given by

v_{ave}\left[t_i,\:t_f\right]=\frac{x\left(t_f\right)-x\left(t_i\right)}{t_f-t_i}

Since the final and initial positions are equal, the average velocity is 0 m/s.

PART B. Find the average velocity over the time interval from 1 to 3 seconds.

ANSWER: 20 m/s 

v_{ave}\left[t_i,\:t_f\right]=\frac{x\left(t_f\right)-x\left(t_i\right)}{t_f-t_i}=\frac{60-20}{3-1}=\frac{40}{2}=20\:m/s

PART C. Now find v_{ave}\left[0,\:3\right].

ANSWER: 13.3 m/s

v_{ave}\left[0,\:3\right]=\frac{x\left(t_f\right)-x\left(t_i\right)}{t_f-t_i}=\frac{60-20}{3-0}=\frac{40}{3}=13.3\:m/s

PART D. Find the average velocity over the time interval from 3 to 6 seconds.

ANSWER: -13.3 m/s

v_{ave}\left[3,\:6\right]=\frac{x\left(t_f\right)-x\left(t_i\right)}{t_f-t_i}=\frac{20-60}{6-3}=\frac{-40}{3}=-13.3\:m/s

PART E. Finally, find the average velocity over the whole time interval shown in the graph.

ANSWER: 0 m/s

Note that though the average velocity is zero for this time interval, the instantaneous velocity (i.e., the slope of the graph) has several different values (positive, negative, zero) during this time interval.

Note as well that since average velocity over a time interval is defined as the change in position (displacement) in the given interval divided by the time, the object can travel a great distance (here 80 meters) and still have zero average velocity, since it ended up exactly where it started. Therefore, zero average velocity does not necessarily mean that the object was standing still the entire time!