Physics, 22.09.2019 21:30 melissa3333

Emmy is standing on a moving sidewalk that moves at +2 m/s. suddenly, she realizes she might miss her flight, so begins to speed up, but then she hears over the pa system that her flight is delayed. emily's acceleration during the 5 s time period in which all of the events occur is shown in the acceleration versus time graph. in the velocity versus time graph, construct a plot of emmy's velocity over this same time period. assuming all the values given are exact, what is emmy's velocity at a time of 3.49s? enter your answer to at least three significant digits.

Emmy is standing on a moving sidewalk that moves at +2 m/s. suddenly, she realizes she might miss he

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Answer from: faithcalhoun

The velocity of Emmy at time of 3.49 s is -8.47 m/s

The given parameters

initial velocity of Emmy, u = 2 m/sfrom the chart, acceleration at time = 3.0 s = -3 m/s²also acceleration at time = 4.0 s = -3 m/s²

The acceleration at time of 3.49 s must be -3 m/s², since the acceleration is constant from 3.0 s to 4.0 s.

The velocity of Emmy at time of 3.49 s is calculated as follows;

v = u + at

where;

a is the acceleration of Emmy at time = 3.49 s

u is the initial velocity of Emmy

v is the final velocity of Emmy at time of 3.49s

$v = 2 \ + \ (-3\times 3.49)\\\\v = 2 - 10.47\\\\v = -8.47 \ m/s$

Thus, the velocity of Emmy at time of 3.49 s is -8.47 m/s

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Emmy is standing on a moving sidewalk that moves at +2 m/s. suddenly, she realizes she might miss he

Answer from: nunezjazmine01

Refer to the attachment for the diagram.

3.53 m/s.

Explanation:

Acceleration is the first derivative of velocity relative to time. In other words, the acceleration is the same as the slope (gradient) of the velocity-time graph. Let represents the time in seconds and the speed in meters-per-second.

For $0 < x \le 1$ :

Initial value of

: $\rm 2\;m\cdot s^{-1}$ at t = 0

; Hence the point on the segment: (0, 2)

. Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 2\; m\cdot s^{-2}$ .Find the equation of this segment in slope-point form: $v - 2 = 2 (t - 0) \implies v = 2t + 2, \quad 0 < t \le 1$ .

Similarly, for $1 < x \le 2$ :

Initial value of

is the same as the final value of

in the previous equation at t = 1

; Hence the point on the segment: (1, 4)

. Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 1\; m\cdot s^{-2}$ .Find the equation of this segment in slope-point form: $v - 4 = (t - 1) \implies v = t + 3 \quad 1 < t \le 2$ .

For $2 < x \le 3$ :

Initial value of

is the same as the final value of

in the previous equation at t = 2

; Hence the point on the segment: (2, 5)

. Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . There's no acceleration. In other words, the velocity is constant.Find the equation of this segment in slope-point form: $v - 5 = 0 (t - 2) \implies v = 5 \quad 2 < t \le 3$ .

For $3 < x \le 4$ :

Initial value of

is the same as the final value of

in the previous equation at t = 3

; Hence the point on the segment: (3, 5)

. Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm -3\; m\cdot s^{-2}$ . In other words, the velocity is decreasing.Find the equation of this segment in slope-point form: $v - 5 = -3 (t - 3) \implies v = -3t + 14 \quad 3 < t \le 4$ .

For $4 < x \le 5$ :

Initial value of

is the same as the final value of

in the previous equation at t = 4

; Hence the point on the segment: (4, 2)

. Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . In other words, the velocity is once again constant.Find the equation of this segment in slope-point form: $v - 2 = 0 (t - 4) \implies v = 2\quad 4 < t \le 5$ .

$t = \rm 3.49\;s$ is in the interval $3 < x \le 4$ . Apply the equation for that interval: $v = -3t +14 = \rm 3.53\; m \cdot s^{-1}$ .

Answer from: Quest

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Answer from: Quest

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