Question

Graph of v 10. (12 Points) A particle move along the x-axis so that the velocity at time, t, for 0 ? t ? 6, is given by a differentiable function v whose graph is shown above. The velocity is 0 at t = 0, t = 3, and t = 5, and the graph has horizontal tangents at t = 1 and t = 4. The areas of the regions bounded by the t-axis and the graph of v on the intervals [0, 3], [3, 5], and [5, 6] are 8, 3, and 2, respectively. At time t = 0, the particle is at x = -2. a) For 0 ? t ? 6, find both the time and the position of the particle when the particle is farthest to the left. Justify your answer. b) For how many values of t, where 0 ? t ? 6, is the particle at x = -8?. Explain your reasoning. c) On the interval 2 < t < 3, is the speed of the particle increasing or decreasing? Give a reason for your answer. d) During what time intervals, if any, is the acceleration of the particle negative? Justify your answer. 

          Graph of v

10. (12 Points) A particle move along the x-axis so that the velocity at time, t, for 0 ? t ? 6, is given by a differentiable function v whose graph is shown above. The velocity is 0 at t = 0, t = 3, and t = 5, and the graph has horizontal tangents at t = 1 and t = 4. The areas of the regions bounded by the t-axis and the graph of v on the intervals [0, 3], [3, 5], and [5, 6] are 8, 3, and 2, respectively. At time t = 0, the particle is at x = -2.

a) For 0 ? t ? 6, find both the time and the position of the particle when the particle is farthest to the left. Justify your answer.

b) For how many values of t, where 0 ? t ? 6, is the particle at x = -8?. Explain your reasoning.

c) On the interval 2 < t < 3, is the speed of the particle increasing or decreasing? Give a reason for your answer.

d) During what time intervals, if any, is the acceleration of the particle negative? Justify your answer.
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Graph of v 10. (12 Points) A particle move along the x-axis so that the velocity at time, t, for 0 ? t ? 6, is given by a differentiable function v whose graph is shown above. The velocity is 0 at t = 0, t = 3, and t = 5, and the graph has horizontal tangents at t = 1 and t = 4. The areas of the regions bounded by the t-axis and the graph of v on the intervals [0, 3], [3, 5], and [5, 6] are 8, 3, and 2, respectively. At time t = 0, the particle is at x = -2. a) For 0 ? t ? 6, find both the time and the position of the particle when the particle is farthest to the left. Justify your answer. b) For how many values of t, where 0 ? t ? 6, is the particle at x = -8?. Explain your reasoning. c) On the interval 2 < t < 3, is the speed of the particle increasing or decreasing? Give a reason for your answer. d) During what time intervals, if any, is the acceleration of the particle negative? Justify your answer.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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10. (12 Points) A particle move along the x-axis so that the velocity at time, t, for 0 ≤ t ≤ 6, is given by a differentiable function v whose graph is shown above. The velocity is 0 at t = 0, t = 3, and t = 5, and the graph has horizontal tangents at t = 1 and t = 4. The areas of the regions bounded by the t-axis and the graph of v on the intervals [0, 3], [3, 5], and [5, 6] are 8, 3, and 2, respectively. At time t = 0, the particle is at x = -2. a) For 0 ≤ t ≤ 6, find both the time and the position of the particle when the particle is farthest to the left. Justify your answer. b) For how many values of t, where 0 ≤ t ≤ 6, is the particle at x = -8?. Explain your reasoning. c) On the interval 2 < t < 3, is the speed of the particle increasing or decreasing? Give a reason for your answer. d) During what time intervals, if any, is the acceleration of the particle negative? Justify your answer.
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Transcript

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00:01 All right, so we have a graph of velocity versus time, and we're told that the areas here, that this one is equal to eight, and then we have this one equal to three, and this one is equal to two.
00:20 And let's see, we're told that the particle is moving along the x -axis, and at time t equals zero, the particle is at x -equal.
00:34 Negative 2.
00:35 So when is the particle for this to the left? so basically we know that the position function is the integration of the velocity function with respect to time, which is equal to the area under the curve.
00:53 So from t equals 0 to 3, we go from negative 2 and then the change in position is going to be negative 8.
01:03 So it's going to be at negative 10.
01:06 And the velocity is negative this whole interval, so it's continually decreasing.
01:11 Then from 3 to 5, we have a change of 3.
01:18 So then the particle goes from negative 10 to plus 3 is negative 7.
01:23 And then from t equals 5 to 6, we have a decrease in position by 2, so negative 9.
01:31 So furthest to the left, meaning the most negative, is right here at negative 10 at time t equals three.
01:40 You only have to consider the end points of the intervals because velocity is either entirely increasing or entirely decreasing in those intervals.
01:51 That's answer choice.
01:53 That's part a...
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