Question

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.

Answer 1

Without a graph or specific numbers for Emmy's acceleration, we cannot compute her exact velocity at 3.49s. Generally, you would integrate the acceleration over time and add the initial velocity to find velocity at a specific time, provided acceleration is known.

Step-by-step explanation:

The question pertains to the calculation of an object's velocity at a specific time when it is undergoing acceleration. Since exact numbers or a graph are not provided in the question, we cannot calculate Emmy's velocity directly. However, generally speaking, to find the velocity of an object at a certain time when its acceleration is known, you would find the area under the acceleration versus time graph up to that point. This area represents the change in velocity. You would then add this change in velocity to the initial velocity to determine the object's velocity at that time.

For example, if the acceleration versus time graph is a straight line, then the change in velocity (Δv) is simply the average acceleration (a) times the time interval (Δt), which can be expressed as Δv = a * Δt. If Emmy's initial velocity is +2 m/s on the moving sidewalk and we assume a constant acceleration, we can use this principle to find her velocity at any given time.

However, without specific data on Emmy's acceleration over time, we cannot provide an exact numeric answer to the question of her velocity at 3.49 seconds.

Answer 2

Refer to the attachment for the diagram.

3.53 m/s.

Step-by-step 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
`t` represents the time in seconds and
`v` the speed in meters-per-second.

For
`0 < x \le 1`:

• Initial value of
  `v`:
  `\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
  `v` is the same as the final value of
  `v` in the previous equation at
  `t = 1`:
  `t = 2t + 2 = 4`; 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
  `v` is the same as the final value of
  `v` in the previous equation at
  `t = 2`:
  `t = t + 3 = 5`; 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
  `v` is the same as the final value of
  `v` in the previous equation at
  `t = 3`:
  `t = 5`; 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
  `v` is the same as the final value of
  `v` in the previous equation at
  `t = 4`:
  `t = -3t + 14`; 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)`.