Physics of Everyday Phenomena
9th Edition
ISBN: 9781259894008
Author: W. Thomas Griffith, Juliet Brosing Professor
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 2, Problem 29CQ
The velocity-versus-time graph of an object curves as shown in the diagram. Is the acceleration of the object constant? Explain.
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Chapter 2 Solutions
Physics of Everyday Phenomena
Ch. 2 - Prob. 1CQCh. 2 - Suppose we choose inches as our basic unit of...Ch. 2 - What units would have an appropriate size for...Ch. 2 - A tortoise and a hare cover the same distance in a...Ch. 2 - A driver states that she was doing 80 when stopped...Ch. 2 - Does the speedometer on a car measure average...Ch. 2 - Is the average speed over several minutes more...Ch. 2 - The highway patrol sometimes uses radar guns to...Ch. 2 - Is the term vehicle density (as used in everyday...Ch. 2 - Prob. 10CQ
Ch. 2 - At the front end of a traffic jam, is the vehicle...Ch. 2 - A hockey puck is sliding on frictionless ice. It...Ch. 2 - A ball attached to a string is whirled in a...Ch. 2 - Prob. 14CQCh. 2 - A dropped ball gains speed as it falls. Can the...Ch. 2 - A driver of a car steps on the brakes, causing the...Ch. 2 - At a given instant in time, two cars are traveling...Ch. 2 - A car just starting up from a stop sign has zero...Ch. 2 - A car traveling with constant speed rounds a curve...Ch. 2 - A racing sports car traveling with a constant...Ch. 2 - In the graph shown here, velocity is plotted as a...Ch. 2 - A car moves along a straight line so that its...Ch. 2 - For the car whose distance is plotted against time...Ch. 2 - A car moves along a straight section of road so...Ch. 2 - For the car whose velocity is plotted in question...Ch. 2 - Look again at the velocity-versus-time graph for...Ch. 2 - Suppose the acceleration of a car increases with...Ch. 2 - When a car accelerates uniformly from rest, which...Ch. 2 - The velocity-versus-time graph of an object curves...Ch. 2 - For a uniformly accelerated car, is the average...Ch. 2 - A car traveling in the forward direction...Ch. 2 - A car starts from rest, accelerates uniformly for...Ch. 2 - Suppose that two runners run a 100-meter dash, but...Ch. 2 - Sketch a graph showing velocity-versus-time curves...Ch. 2 - A physics instructor walks with increasing speed...Ch. 2 - Prob. 36CQCh. 2 - Return to example box 2.4, but this time assume...Ch. 2 - A traveler covers a distance of 413 miles in a...Ch. 2 - A walker covers a distance of 2.4 km in a time of...Ch. 2 - Grass clippings are found to have an average...Ch. 2 - A driver drives for 2.5 hours at an average speed...Ch. 2 - A woman walks a distance of 504 m, with an average...Ch. 2 - A person in a hurry averages 70 MPH on a trip...Ch. 2 - A hiker walks with an average speed of 1.3 m/s....Ch. 2 - Prob. 8ECh. 2 - A car travels with an average speed of 65 MPH....Ch. 2 - Starting from rest and moving in a straight line,...Ch. 2 - Starting from rest, a car accelerates at a rate of...Ch. 2 - The velocity of a car decreases from 28 m/s to 20...Ch. 2 - A car traveling with an initial velocity of 16 m/s...Ch. 2 - A runner traveling with an initial velocity of 1.1...Ch. 2 - A car moving with an initial velocity of 32 m/s...Ch. 2 - A runner moving with an initial velocity of 4.0...Ch. 2 - If a world-class sprinter ran a distance of 100...Ch. 2 - Starting from rest, a car accelerates at a...Ch. 2 - A railroad engine moves forward along a straight...Ch. 2 - The velocity of a car increases with time, as...Ch. 2 - A car traveling due west on a straight road...Ch. 2 - A car traveling in a straight line with an initial...Ch. 2 - Just as car A is starting up, it is passed by car...
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- A student drives a moped along a straight road as described by the velocity-versus-time graph in Figure P2.12. Sketch this graph in the middle of a sheet of graph paper. (a) Directly above your graph, sketch a graph of the position versus time, aligning the time coordinates of the two graphs. (b) Sketch a graph of the acceleration versus time directly below the velocity-versus-time graph, again aligning the time coordinates. On each graph, show the numerical values of x and ax for all points of inflection. (c) What is the acceleration at t = 6.00 s? (d) Find the position (relative to the starting point) at t = 6.00 s. (e) What is the mopeds final position at t = 9.00 s? Figure P2.12arrow_forwardA speedboat travels in a straight line and increases in speed uniformly from i = 20.0 m/s to f = 30.0 m/s in displacement x of 200 m. We wish to find the time interval required for the boat to move through this displacement, (a) Draw a coordinate system for this situation, (b) What analysis model is most appropriate for describing this situation? (c) From the analysis model, what equation is most appropriate for finding the acceleration of the speedboat? (d) Solve the equation selected in part (c) symbolically for the boats acceleration in terms of i, f, and x. (e) Substitute numerical values lo obtain the acceleration numerically. (f) Find the time interval mentioned above.arrow_forwardThe Acela is an electric train on the WashingtonNew YorkBoston run, carrying passengers at 170 mi/h. A velocitytime graph for the Acela is shown in Figure P2.46. (a) Describe the trains motion in each successive time interval. (b) Find the trains peak positive acceleration in the motion graphed. (c) Find the trains displacement in miles between t = 0 and t = 200 s. Figure P2.46 Velocity versus time graph for the Acela.arrow_forward
- The Acela is an electric train on the Washington-New YorkBoston run, carrying passengers at 170 mi/h. A velocity-time graph for the Acela is shown in Figure P2.69. (a) Describe the train's motion in each successive lime interval, (b) Find the trains peak positive acceleration in the motion graphed, (c) Find the trains displacement in miles between t = 0 and t = 200 s.arrow_forwardDraw motion diagrams for (a) an object moving to the right at constant speed, (b) an object moving to the right and speeding up at a constant rate, (c) an object moving to the right and slowing down at a constant rate, (d) an object moving to the left and speeding up at a constant rate, and (e) an object moving to the left and slowing down at a constant rate. (f) How would your drawings change if the changes in speed were not uniform, that is, if the speed were not changing at a constant rate?arrow_forwardAn object is at x = 0 at t = 0 and moves along the x axis according to the velocitytime graph in Figure P2.40. (a) What is the objects acceleration between 0 and 4.0 s? (b) What is the objects acceleration between 4.0 s and 9.0 s? (c) What is the objects acceleration between 13.0 s and 18.0 s? (d) At what time(s) is the object moving with the lowest speed? (e) At what time is the object farthest from x = 0? (f) What is the final position x of the object at t = 18.0 s? (g) Through what total distance has the object moved between t = 0 and t = 18.0 s? Figure P2.40arrow_forward
- PROBLEM A race car starting from rest accelerates at a constant rate of 5.00 m/s2, (a) What is the velocity of the car after it has traveled 1.00 102 ft? (b) How much time has elapsed? (c) Calculate the average velocity two different ways. STRATEGY Weve read the problem, drawn the diagram in Figure 2.16, and chosen a coordinate system (steps 1 and 2). We'd like to find the velocity v after a certain known displacement x. The acceleration a is also known, as is the initial velocity v0 (step 3, labeling, is complete), so the third equation in Table 2.4 looks most useful for solving part (a). Given the velocity, the first equation in Table 2.4 can then be used to find the time in part (b). Part (c) requires substitution into Equations 2.2 and 2.7, respectively. Figure 2.16 (Example 2.4) SOLUTION (a) Convert units of x to SI, using the information in the inside front cover. Write the kinematics equation for v2 (step 4): Solve for v, taking the positive square root because the car moves to the right (step 5): Substitute v0 = 0, a = 5.00 m/s2, and x = 30.5 m: 1.00 102ft = (1.00 102 ft) v2 = v02 + 2a x v = v02+2ax v = v02+2ax = (0)2+2(5.00m/s2)(30.5m)= 17.5 m/s (b) Find the trooper's speed at that time. Substitute the time into the troopers velocity equation: vtrooper = v0 + atrooper t = 0 + (3.00m/s2)(16.9s) = 50.7 m/s Solve Example 2.5, Car Chase, by a graphical method. On the same graph, plot position versus time for the car and the trooper. From the intersection of the two curves, read the time at which the trooper overtakes the car.arrow_forward(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?arrow_forwardA motorist drives for 35.0 minutes at 85.0 km/h and then stops for 15.0 minutes. He then continues north, traveling 130. Km in 2.00 h. (a) What is his total displacement? (b) What is his average velocity?arrow_forward
- Plus and minus signs are used in one-dimensional motion to indicate direction. What is the sign of an acceleration that reduces the magnitude of a negative velocity? Of a positive velocity?arrow_forwardAn object that moves in one dimension has the velocity-versus-time graph shown in Figure P2.52. At time t = 0, the object has position x = 0. a. At time t = 5 s. is the acceleration of the object positive, negative, or zero? Explain. b. At time t = 8 s, is the object speeding up, showing down, or moving with constant speed? Explain. c. Write an expression for the position of the object as a function of time. Explain how you use the graph to obtain your answer. d. Use your expression from part (c) to determine the time (if any) at which the object reaches its maximum position. Check your results by examining the graph. Hint: To get started with finding the maximum of a function, take the derivative and set it equal to zero.arrow_forwardFigure below is a velocity-versus-time graph for a moving object. Match the answers with questions for the object's velocity (v) and acceleration (a) at different time instants. A time in seconds v att = 1.0 s A. V + 0; a = 0 v att = 2.0s B. V + 0; a +0 v att = 4.0 s C. v = 0; a = 0 D.v = 0; a +0 velocity in meters/secondarrow_forward
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Position/Velocity/Acceleration Part 1: Definitions; Author: Professor Dave explains;https://www.youtube.com/watch?v=4dCrkp8qgLU;License: Standard YouTube License, CC-BY