(a) Sketch a graph of acceleration versus time corresponding to the graph of velocity versus time given in Figure 2.57. (b) Identify the time or tim ? es ( ?ta , ? ? tb , t? c , etc.) at which the acceleration is greatest. (c) At which times is it zero? (d) At which times is it negative?
Step-by-step solution 29CQ Step 1 of 8 (a) The slope of the graph of velocity versus time gives the acceleration. If tangent lines are drawn on the given points then the tangents shows the direction of acceleration: Step 2 of 8 The slope is constant at b and c and zero at d, therefore the velocity start at some positive value, remain constant for some time then become zero at d. The slope at e is zero, so acceleration increased thendecreased and it becomes zero at e, again the slope is constant at and g then zero at h, slope is negative at i , then constant at j, k and l Step 3 of 8 The graph for acceleration and time is as below: Step 4 of 8 The graph is drawn by the help of velocity time graph and the conclusion of the velocity time graph written above.
Textbook: Physics: Principles with Applications
Author: Douglas C. Giancoli
This textbook survival guide was created for the textbook: Physics: Principles with Applications, edition: 6. This full solution covers the following key subjects: acceleration, versus, times, graph, identify. This expansive textbook survival guide covers 35 chapters, and 3914 solutions. Physics: Principles with Applications was written by and is associated to the ISBN: 9780130606204. The full step-by-step solution to problem: 29CQ from chapter: 2 was answered by , our top Physics solution expert on 03/03/17, 03:53PM. The answer to “(a) Sketch a graph of acceleration versus time corresponding to the graph of velocity versus time given in Figure 2.57. (b) Identify the time or tim ? es ( ?ta , ? ? tb , t? c , etc.) at which the acceleration is greatest. (c) At which times is it zero? (d) At which times is it negative?” is broken down into a number of easy to follow steps, and 59 words. Since the solution to 29CQ from 2 chapter was answered, more than 284 students have viewed the full step-by-step answer.