graphs-of-the-velocity-functions-of-two-particles-are-shown-where-t-is-measured-in-seconds-when-is-e

Name: graphs of the velocity functions of two particles are shown where t is measured in seconds when is e
Uploaded: 2020-05-13
Duration: 3 min 34 s
Channel: Numerade
Description: Graphs of the velocity functions of two particles are shown, where $t$ is measured in seconds. When is each particle speeding up? When is it slowing down? Explain.

Question

Graphs of the velocity functions of two particles are shown, where $t$ is measured in seconds. When is each particle speeding up? When is it slowing down? Explain.

Heather Zimmers · Accepted Answer

here we have the graph of velocity function, and we&#x27;re going to figure out when the particle is speeding up and when it&#x27;s slowing down. So what we need to know is that a particle is speeding up when the velocity and the acceleration have the same sign. So that means they could be both positive or both negative. And the particle is slowing down when the velocity and the acceleration have different signs. So one positive one negative. So let&#x27;s take a look at the graph and let&#x27;s break it down into the changing, um, intervals. So it looks like it&#x27;s doing something on the interval from 0 to 1. And then it&#x27;s doing something different on the interval from 1 to 2. And then it&#x27;s doing something different on the interval from 2 to 3. So let&#x27;s analyze the velocity and the acceleration on each of those intervals. Okay, so when the graph is above the X axis, the velocity is positive. So we see that for the first interval, the velocities positive and for the second interval of last year&#x27;s positive, When the graph is below the X axis, the velocity is negative. So from 2 to 3 of the velocity is negative. When the velocity graph is increasing, the acceleration is positive. So we can see the velocity graph increasing from 0 to 1. And when the velocity graph is decreasing, the acceleration is negative. So we can see that happening from 1 to 2 and from 2 to 3. Okay, Notice that we have the same sign from time zero to time one and the same sign from time 32 time from time to time. Three. So that means that the particle is speeding up for those times speeding up 01 and 2 to 3 and then notice that we have different signs on the interval from 1 to 2. So the particle is slowing down from the on the interval from 1 to 2. Okay, we can do part B the same way. And we&#x27;re interested in times from 0 to 4. So let&#x27;s break this down. Something&#x27;s happening from time 01 Something else is happening from 1 to 2. Something else is happening from 2 to 3, and something else is happening from 3 to 4. Now we want to take a look at what&#x27;s happening to the velocity and what&#x27;s happening to the acceleration for each interval. So the velocity is positive from time zero to time. Three because the graph is above the X axis. So we have positive, positive, positive, negative, and the acceleration is positive whenever the velocity is increasing, so that would be from 1 to 2 and the velocity. The acceleration is negative whenever the velocity is decreasing. So that would be from 0 to 1 from 2 to 3 and from 3 to 4. All right, so let&#x27;s look for where we have the same sign. We have the same sign from 1 to 2 and 3 to 4, so that means it&#x27;s speeding up on those intervals and we have opposite signs from 0 to 1 and from 2 to 3, so it&#x27;s slowing down on those intervals.

Problem

Graphs of the $ position $ functions of two parti…

Problem 5 Hard Difficulty

Graphs of the velocity functions of two particles are shown, where $t$ is measured in seconds. When is each particle speeding up? When is it slowing down? Explain.

Answer

a) Speeding up: $0 < t < 1$ and $2 < t <3$
Slowing down: $1 < t < 2$
b) speeding up: $1 < t < 2 a n d 3 < t < 4$
slowing down: $0 < t < 1$ and $2 < t < 3$

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a) Speeding up: $0 < t < 1$ and $2 < t <3$Slowing down: $1 < t < 2$b) speeding up: $1 < t < 2 a n d 3 < t < 4$slowing down: $0 < t < 1$ and $2 < t < 3$

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Michael J.

Joseph L.

a) Speeding up: $0 < t < 1$ and $2 < t <3$
Slowing down: $1 < t < 2$
b) speeding up: $1 < t < 2 a n d 3 < t < 4$
slowing down: $0 < t < 1$ and $2 < t < 3$