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.

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 [tex]t[/tex] represents the time in seconds and [tex]v[/tex] the speed in meters-per-second.

For [tex]0 < x \le 1[/tex]:

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

Similarly, for [tex]1 < x \le 2[/tex]:

Initial value of [tex]v[/tex] is the same as the final value of [tex]v[/tex] in the previous equation at [tex]t = 1[/tex]: [tex]t = 2t + 2 = 4[/tex]; Hence the point on the segment: [tex](1, 4)[/tex].
Slope of the velocity-time graph is the same as acceleration during that period of time: [tex]\rm 1\; m\cdot s^{-2}[/tex].

Find the equation of this segment in slope-point form: [tex]v - 4 = (t - 1) \implies v = t + 3 \quad 1 < t \le 2[/tex].

For [tex]2 < x \le 3[/tex]:

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

For [tex]3 < x \le 4[/tex]:

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

For [tex]4 < x \le 5[/tex]:

Initial value of [tex]v[/tex] is the same as the final value of [tex]v[/tex] in the previous equation at [tex]t = 4[/tex]: [tex]t = -3t + 14[/tex]; Hence the point on the segment: [tex](4, 2)[/tex].
Slope of the velocity-time graph is the same as acceleration during that period of time: [tex]\rm 0\; m\cdot s^{-2}[/tex]. In other words, the velocity is once again constant.
Find the equation of this segment in slope-point form: [tex]v - 2 = 0 (t - 4) \implies v = 2\quad 4 < t \le 5[/tex].

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

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-09-09T16:03:47+02:00

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/s

from the chart, acceleration at time = 3.0 s = -3 m/s²

also 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;

[tex]v = u + at[/tex]

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

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

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

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