. . V P rt A Two objects start at the same place at the same time a and move along the same straight line. The figure below shows the position x as a function of time t for each object. (gure 1) At point A, what must be true about the motion of these objects? Select all that apply. Both have the same speed. Both are at the same position. Both have the same velocity. Figure h Both have traveled the same distance. ue-st Answer I Provide Feedback Next > 7 An object starts from rest and accelerates Part A uniformly. If it moves 2.00 m during the first second, then, during the first 5.00 seconds it will move 7.00 m. 10.0 m. 25.0 m. 50.0 m. Which figure could represent the velocity versus time graph of a motorcycle whose speed is increasing? v (m/s) v (m/s) O TTTTT I (S) v (m/s) v (m/s) TITTTI (S )The gure shows the velocity versus time graph for a car driving on a straight road. Which of the following best describes the acceleration of the car? v(nvs) r(s) The acceleration of the car is negative and decreasing. The acceleration of the car is constant. The acceleration oi the car is positive and decreasing. The acceleration of the car is negative and increasing. The acceleration of the car is positive and increasing. A nerve conduction velocity test (NOV) is an ' Part A electrical test that is used to evaluate the function of nerves. A NCV on a patient's sciatic _ _ nerve (length 010-80 In} reveals a conduction How long did It take the impulse to travel along the nerve? velocity ot 10 m/s. Learning Goal: Part A Figure 1 In this problem you will determine the average velocity of a moving object from the graph of Consulting the graph shown in the figure, find the object's average velocity over the time interval from 0 to 1 its position x(t) as a function of time t. A second. raveling object might move at different speeds Answer to the nearest integer. and in different directions during an interval of time, but if we ask at what constant velocity View Available Hint(s) the object would have to travel to achieve the same displacement over the given time interval, that is what we call the object' Templates Symbols undo redo reset keyboard shortcuts help, average velocity. We will use the notation vavet1, t2 to indicate average velocity over the vave0, 1] = m/s time interval from t1 to t2. For instance, vavel, 3] is the average velocity over the time interval from t = 1 to t = 3. Submit Part B Find the average velocity over the time interval from 1 to 3 seconds. Express your answer in meters per second to the nearest integer. View Available Hint(s) Templates Symbols undo redo reset keyboard shortcuts help, vavel, 3] = m/s Submit Part C Now find vave0, 3]. Give your answer to three significant figures. Figure View Available Hint(s) 60- Templates Symbols undo redo reset keyboard shortcuts help, 40- vave0, 3] = m/s x (meters) 20 Submit 2 5 6 t (seconds)Learning Goal: Part A Figure 1 In this problem you will determine the average velocity of a moving object from the graph of Part B its position x(t) as a function of time t. A raveling object might move at different speeds and in different directions during an interval of Part C time, but if we ask at what constant velocity the object would have to travel to achieve the same displacement over the given time Part D interval, that is what we call the object's average velocity. We will use the notation vavet1, t2 to indicate average velocity over the Find the average velocity over the time interval from 3 to 6 seconds. time interval from t1 to t2. For instance, vavel, 3] is the average velocity over the time Express your answer to three significant figures. interval from t = 1 to t = 3. > View Available Hint(s) Templates Symbols undo redo reset keyboard shortcuts help, Vave3.0, 6.0] = m/s Submit Part E Finally, find the average velocity over the whole time interval shown in the graph. Express your answer to three significant figures. View Available Hint(s) Templates Symbols undo redo reset keyboard shortcuts help, vave0.0, 6.0] = m/s Figure Submit Provide Feedback Next > 60- 40- x (meters) 20 2 5 6 t (seconds)To describe the motion of a particle along a Now let's study the graph shown in the figure in more detail. (Figure 1) Refer to this graph to answer Parts A, B, and straight line, it is often convenient to draw a C. graph representing the position of the particle at different times. This type of graph is usually referred to as an x vs. t graph. To draw such a Part A graph, choose an axis system in which time t is plotted on the horizontal axis and position z on the vertical axis. Then, indicate the values of x at various times t. Mathematically, this What is the overall displacement Ax of the particle? corresponds to plotting the variable r as a Express your answer in meters. function of t. An example of a graph of position as a function of time for a particle View Available Hint(s) traveling along a straight line is shown below. Note that an z vs. t graph like this does not represent the path of the particle in space. Templates Symbols undo redo reset keyboard shortcuts help, Ax m Submit Part B What is the average velocity vavof the particle over the time interval At = 50.0 s ? Express your answer in meters per second. View Available Hint(s) emplates Symbols undo redo reset keyboard shortcuts help, Vav m/s Submit Part C Figure What is the instantaneous velocity v of the particle at t = 10.0 s? x (m) 40 Express your answer in meters per second. View Available Hint(s) 30 20 templates Symbols undo redo reset keyboard shortcuts help, 10 m/s 0 10 20 30 40 50* * (s )To describe the motion of a particle along a Part C straight line, it is often convenient to draw a graph representing the position of the particle at different times. This type of graph is usually Another common graphical representation of motion along a straight line is the v vs. t graph, that is, the graph of referred to as an a vs. t graph. To draw such a (instantaneous) velocity as a function of time. In this graph, time t is plotted on the horizontal axis and velocity v on graph, choose an axis system in which time t the vertical axis. Note that by definition, velocity and acceleration are vector quantities. In straight-line motion, is plotted on the horizontal axis and position r however, these vectors have only one nonzero component in the direction of motion. Thus, in this problem, we will call on the vertical axis. Then, indicate the values the velocity and a the acceleration, even though they are really the components of the velocity and acceleration of x at various times t. Mathematically, this vectors in the direction of motion. corresponds to plotting the variable r as a function of t. An example of a graph of position as a function of time for a particle Part D traveling along a straight line is shown below. Note that an z vs. t graph like this does not represent the path of the particle in space. Which of the graphs shown is the correct v vs. t plot for the motion described in the previous parts? 4 B (m/s) u (m/s) 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0 10 20 30 40 50 0 10 20 30 40 50* (5) D U (m/s) u (m/s) 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0 10 20 30 40 50 0 10 20 30 40 50' ($) View Available Hint(s) Graph A O Graph B Figure Graph C Graph D x (m) 40 Submit Request Answer 30 20 Part E Complete previous part(s) 10 Provide Feedback 0 Next > 10 20 30 40 50* ($ )Part A Match each graph of velocity as a function of time to its respective qualitative graph of acceleration as a function of time. Drag the appropriate labels to their respective targets. Reset Help a x a x a x a x o o O U x Ux Ux OWhen you normally drive the freeway between Part A Sacramento and San Francisco at an average speed of 115 km/h (71.5 mi/h), the trip takes 1 h and 13 min. On a Friday afternoon, How much longer does the trip take on Friday than on the other days? however, heavy traffic slows you down to an Express your answer in minutes. average of 80.0 km/h (49.7 mi/h) for the same distance. Templates Symbols undo redo reset keyboard shortcuts help, At = minThe driver of a car traveling on the highway suddenly slams on the brakes because of a slowdown in traffic ahead. Part A If the car's speed decreases at a constant rate from 60 mi/h to 50 mi/h in 3.0 s, what is the magnitude of its acceleration, assuming that it continues to move in a straight line? Express your answer in miles per hour squared. Templates Symbols undo redo reset keyboard shortcuts help, mi/h2 Submit Request Answer Part B What distance does the car travel during the braking period? Express your answer in feet. Templates Symbols undo redo reset keyboard shortcuts help, d = ft Submit Request AnswerA brick is released with no initial speed from the roof of a building and strikes the ground in 1 .60 s , encountering no appreciable air drag. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution oi A ball on the roof. v PartA How tall, in meters, is the building? Express your answer In meters. v L it ' - - undo redo re Keyboard shortcuts gall)? Quest Answer v Part3 How fast is the brick moving just before it reaches the ground? Express your answer In meters per second. nest Answer Provide Feedback Next) Now let's examine the case of a ball being thrown vertically upward. Even though the ball is initially moving upward, it is, nevertheless, in free fall and we can use the following equations to analyze its motion. \"Oy'gt _ 1 2 y , yo+voiggt v32; 031,7 290.- 7 110) Suppose you throw a ball vertically upward from the flat roof of a tall building. The ball leaves your hand at a point even with the roof railing, with an upward velocity of 15.0 m/s. On its way back down, it just misses the railing. Find (a) the position and velocity of the ball 1.00 s and 4.00 s after it leaves your hand; (I!) the velocity of the ball when it is 5.00 m above the railing; and (c) the maximum height reached and the time at which it is reached. Ignore the eflects of the air. Figure 1 of 2 > Theblll actually mm straight up and Y dwasuaight down, we Show ...._M_ aU-shlped pad] fordmilyh? : " SOLUHON SET UP As shown in (gure 1), we place the origin at the level of the roof railing, where the ball leaves your hand, and we take the positive direction to be upward. Here's what we know: The initial position yo is zero, the initial velocity voy(y component) is +15.0 m/s,and the acceleration (y component) is y: 9.80 m/az. SOLVE What equations do we have to work with? The velocity 03, at any time t is uy= ayl ayt = 15.0 m/s + (9.30 nut/52}: The position y at any time t is y = mg: + nytz = (15.0 net/s): + %(79.80 m/52)t2 The velocity \"y at any position y is given by v5 120%? + 26M? '90) (15.0 mm? + 2(9.80 m/SZXB' - 0) Part (a): When i = 1.00 s, the first two equations give y = +10.1 m, y: +5.20 rules. The ball is 10.1 m above the origin (y is positive), and it has an upward velocity (v is positive) of 5.20 m/s (less than the initial velocity of 15.0 m/s, as expected). When 1 = 4.00 a, the same equations give y = 18.4 to, try: 24.2 m/s Thus, at time t = 4.00 s, the ball has passed its highest point and is 18.4 m belowthe origin (y is negative). It has a downward velocity (vy is negative) with magnitude 24.2 m/s (greater than the initial velocity, as we should expect). Note that, to get these results, we don't need to find the highest point the ball reaches or the time at which it is reached. The equations of motion give the position and velocity at anytime, whether the ball is on the way up or on the way down. Part (b): When the ball is 5.00 m above the origin, 3; = +5.00 m. Now we use our third equation to nd the velocity \"y at this point: \"f, (15.0 uni/st)z + 2(4180 m/sz)(5.00 m) 127 1112/52, \"y. = :l:ll.3 m/a We get two values of 113., one positive and one negative. That is, the ball passes the point y = +5.00 in twice, once on the way up and again on the way down. The velocity on the way up is +11.3 m/s, and on the way down it is , 11.3 m/s. Part (c): At the highest point, the ball's velocity is momentarily zero (0y = 0); it has been going up (positive 0y) and is about to start going down (negative uy). From our third equation, we have 0 = (15.0 mm! (19.5 m/y and the maximum height (where y: 0) is y = 11.5 m. We can now find the time 1 when the ball reaches its highest point from the equation vy: 1103.7 gt, setting 11y = 0: _ 15.0 m/s + (9.8 m/az): t = 1.53 3 Alternative Solution: Alternatively, to nd the maximum height, we may ask rst when the maximum height is '..... .- -- .n- - -. an.\" point from the equation vy = voy- gt, setting vy = 0: Now let's examine the case of a ball being thrown vertically upward. Even though the ball is initially moving upward, it is, nevertheless, in = 15.0 m/s + (-9.8 m/s2)t free fall and we can use the following = 1.53 s equations to analyze its motion. Alternative Solution: Alternatively, to find the maximum height, we may ask first when the maximum height is reached. That is, at what value of t is vy = 0? As we just found, vy = 0 when t = 1.53 s. Substituting this value of t vy voy - gt back into the equation for y, we find that y yo + voyt - 59t 2 voy - 29(y - yo) y = (15 m/s)(1.53 s) + 7(-9.8 m/$2) (1.53 s) What velocity is the ball traveling at when it hits the ground? y (m) By (m/s) 15 15 .. The acceleration is Express your answer in meters per second. 10 10 constant (straight line) and negative (negative slope). Templates Symbols undo redo reset keyboard shortcuts help, -5 -51 -10 10 -15 m/s -15 -20 -20 -25 Submit (a) (b) Request Answer (a) Position and (b) velocity as functions of time for a ball thrown upward with an initial velocity of 15 m/s Provide Feedback Next >Part A A car sitting at a red light begins to accelerate at 2.0 m/s when the light turns green. It continues with this acceleration until it reaches a speed of 20 m/s. It then travels at this speed for another few minutes. How far does the car travel in the first 36 s after the light changes to green? Express your answer in meters. Templates Symbols undo redo reset keyboard shortcuts help, m