(b) Calculate the average velocity (in m/s) during these two time intervals.SOLUTIONIn the first time interval, use the equation for average velocity with At = t, - t; = tp - t, = 1 s:ДхA - B =Vx, avg (A → B)m/sAtIn the second time interval, At = 2 s:AXB - Dх, avg (В - D)m/sAtThese values are the same as the ---Select---of the lines joining these points in the figure.(c) Find the instantaneous velocity of the particle (in m/s) at t = 2.5 s.SOLUTIONMeasure the slope of the green line at t = 2.5 s (point C) in the figure:m/sNotice that this instantaneous velocity is on the same order of magnitude as our previous results, that is, a few meters per second. Is that what you would have expected?EXERCISESuppose a particle moves along the x-axis. Its x-coordinate varies with time according to the expression x = 34.0 + 4.0t + 1.2t“, where x is in meters and t is in seconds.Hint(a) Determine the displacement Ax of the particle (in m) in the time interval t = 9 s to t = 20 s.Ax =m(b) Calculate the average velocity (in m/s) in the time interval t = 9 s to t = 20 s.m/s(c) Find the instantaneous velocity of the particle (in m/s) at t = 18.4 s.V =m/s Average and Instantaneous VelocityA particle moves along the x-axis. Its position varies with time according to the expression x = -4t + 2t2, where x is in meters and t is in seconds. The position-time graph for this motion is shown in the figure below.x (m)108.Slope6.= +4 m/sSlope= -2 m/s4At (s)-2B.-4134Because the position of the particle is given by a mathematical function, the motion of the particle is completely known. Notice that the particle moves in the negative x-direction for the first second of motion, is momentarily at rest at the moment t = 1 s, and moves inthe positive x-direction at times t > 1 s. (For the following questions, assume that the positive direction is to the right. Indicate the direction with the signs of your answers.)(a) Determine the displacement of the particle (in m) in the time intervals t = 0 to t = 1 s and t = 1 s to t = 3 s.SOLUTIONFrom the graph, form a mental representation of the motion of the particle. Keep in mind that the particle does not move in a curved path in space such as that shown by the brown curve in the graphical representation. The particle moves only along the x-axis inone dimension. At t = 0, the particle is moving to the ---Select--- VDuring the first time interval, the slope is negative and hence the average velocity is negative. Therefore, we know that the displacement between A and B must be aand D to be ---Select--- V-Select--- V number having units of meters. Similarly, we expect the displacement between BIn the first time interval, set t= t, = 0 andA= t, = 1s and use the following equation to find the displacement:AXA - B= X- XXBAAXA - Bm=For the second time interval (t = 1 s tot = 3 s), set t, = tp = 1 s and t, = tp = 3 s:AxBAxB - DXf - X; = XD – XB=These displacements can also be read directly from the-Select---V graph.

Answered: Image /qna-images/answer/b3d17a9f-ffbe-4463-89a5-10dfe7e2fa4e.jpg

SOLUTION In the first time interval, use the equation for average velocity with At = t, - t, = te - ta = 1 s: Дх, = x, avg (A - B) A -B = m/s At In the second time interval, At = 2 s: AXB -D = Vx, avg (B - D) m/s At These values are the same as the --Select-- V of the lines joining these points in the figure. (c) Find the instantaneous velocity of the particle (in m/s) at t = 2.5 s. SOLUTION Measure the slope of the green line at t = 2.5 s (point C) in the figure: Vy = m/s Notice that this instantaneous velocity is on the same order of magnitude as our previous results, that is, a few meters per second. Is that what you would have expected? EXERCISE Suppose a particle moves along the x-axis. Its x-coordinate varies with time according to the expression x = 34.0 + 4.0t + 1.2t, where x is in meters and t is in seconds. Hint (a) Determine the displacement Ax of the particle (in m) in the time interval t = 9 s to t = 20 s. Ax = m (b) Calculate the average velocity (in m/s) in the time interval t = 9 s to t = 20 s. m/s (c) Find the instantaneous velocity of the particle (in m/s) att = 18.4 s. V = m/s

SOLUTION In the first time interval, use the equation for average velocity with At = t, - t, = te - ta = 1 s: Дх, = x, avg (A - B) A -B = m/s At In the second time interval, At = 2 s: AXB -D = Vx, avg (B - D) m/s At These values are the same as the --Select-- V of the lines joining these points in the figure. (c) Find the instantaneous velocity of the particle (in m/s) at t = 2.5 s. SOLUTION Measure the slope of the green line at t = 2.5 s (point C) in the figure: Vy = m/s Notice that this instantaneous velocity is on the same order of magnitude as our previous results, that is, a few meters per second. Is that what you would have expected? EXERCISE Suppose a particle moves along the x-axis. Its x-coordinate varies with time according to the expression x = 34.0 + 4.0t + 1.2t, where x is in meters and t is in seconds. Hint (a) Determine the displacement Ax of the particle (in m) in the time interval t = 9 s to t = 20 s. Ax = m (b) Calculate the average velocity (in m/s) in the time interval t = 9 s to t = 20 s. m/s (c) Find the instantaneous velocity of the particle (in m/s) att = 18.4 s. V = m/s

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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