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

Answer:

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 $t$ 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: $v = -3t +14 = \rm 3.53\; m \cdot s^{-1}$.