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From \(t=0\) to \(t=5.00 \mathrm{~min}\), a man stands still, and from \(t=5.00 \mathrm{~min}\) to \(t=10.0 \mathrm{~min}\), he walks briskly in a straight line at a constant speed of \(2.20 \mathrm{~m} / \mathrm{s}\). What are (a) his average velocity \(v_{\text {avg }}\) and (b) his average acceleration \(a_{\text {avg }}\) in the time interval \(2.00\) min to \(8.00 \mathrm{~min} ?\) What are (c) \(v_{\text {avg }}\) and (d) \(a_{\text {avg }}\) in the time interval \(3.00 \mathrm{~min}\) to \(9.00 \mathrm{~min} ?\) (e) Sketch \(x\) versus \(t\) and \(v\) versus \(t\), and indicate how the answers to (a) through (d) can be obtained from the graphs.

Short Answer

Expert verified
(a) 1.10 m/s, (b) 0.0061 m/s², (c) 1.47 m/s, (d) 0.0061 m/s².

Step by step solution

01

Understanding the Problem

We have two phases: from 0 to 5 minutes the man is still (velocity = 0), and from 5 to 10 minutes he walks at 2.20 m/s. We need average velocities and accelerations for different intervals.
02

Convert Time to Seconds

Convert the time interval to seconds as it helps in calculation: - 2.00 minutes = 120 seconds - 8.00 minutes = 480 seconds - 3.00 minutes = 180 seconds - 9.00 minutes = 540 seconds.
03

Calculate Distance Travelled

From 5.00 minutes (300 seconds) to when the man is moving (up to 8.00 and 9.00 minutes), we calculate the distance:- From 5.00 to 8.00 minutes (300s to 480s):\[\text{Distance} = (480 - 300) \times 2.20 = 396 \, \text{meters}\]- From 5.00 to 9.00 minutes (300s to 540s):\[\text{Distance} = (540 - 300) \times 2.20 = 528 \, \text{meters}\]
04

Calculate Average Velocity for Intervals

(a) For 2 min to 8 min:- Time period: 480 - 120 = 360 seconds- Total distance: 396 meters (only moving from 300s to 480s)\[v_{\text{avg}} = \frac{396 \text{ m}}{360 \text{ s}} = 1.10 \text{ m/s}\](c) For 3 min to 9 min:- Time period: 540 - 180 = 360 seconds- Total distance: 528 meters (only moving from 300s to 540s)\[v_{\text{avg}} = \frac{528 \text{ m}}{360 \text{ s}} = 1.47 \text{ m/s}\]
05

Calculate Average Acceleration for Intervals

(b) For 2 min to 8 min, the velocity changes from 0 to 2.20 m/s at 300s and remains constant to the end:- \(a_{\text{avg}} = \frac{2.20 \text{ m/s} - 0}{360 \text{ s}} \approx 0.0061 \text{ m/s}^2\)(d) For 3 min to 9 min, the same velocity change occurs:- \(a_{\text{avg}} = \frac{2.20 \text{ m/s} - 0}{360 \text{ s}} \approx 0.0061 \text{ m/s}^2\)
06

Sketch the Graphs

(e) For an \(x\) vs. \(t\) graph, plot a line at \(x = 0\) for 0 to 300 seconds, then a linear increase from 300 to 600 seconds.For a \(v\) vs. \(t\) graph, plot at \(v = 0\) m/s from 0 to 300 seconds, then a flat line at \(v = 2.20\) m/s from 300 to 600 seconds.Average velocities in parts (a) and (c) are depicted by the slope of the \(x\) vs. \(t\) line during the respected intervals, while average accelerations in parts (b) and (d) are depicted by any changes in slope on the \(v\) vs. \(t\) graph.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Average Velocity
To understand average velocity, consider it as the total displacement divided by the total time taken. It takes into account all the movements and pauses experienced in the time interval.
In our exercise, the man initially stands still for 5 minutes, then moves at a constant speed for the next 5 minutes. This means his motion involves a phase of no movement and a phase of consistent motion.

For the time interval between 2 minutes (120 seconds) and 8 minutes (480 seconds), he begins moving at the 5-minute mark (300 seconds). Only this movement is considered for calculating average velocity.
The formula used is:
\[v_{\text{avg}} = \frac{\text{Total Distance Traveled}}{\text{Total Time}}\]
- For the interval from 2 min to 8 min, he covers 396 meters in 360 seconds, giving an average velocity of 1.10 m/s.
- For the interval from 3 min to 9 min, he covers 528 meters also in 360 seconds, resulting in a higher average velocity of 1.47 m/s.

This analysis shows how changes in movement affect average velocity over different periods.
Average Acceleration
Average acceleration is the change in velocity over a period of time. It tells us how quickly an object's velocity changes. In our scenario, the man transitions from standing still to walking at 2.20 m/s.
The average acceleration can be calculated keeping in mind the time intervals in which the change occurred. Since velocity increases from 0 to 2.20 m/s only when he starts walking, only the duration of his walk impacts the calculation.

The formula for average acceleration is:
\[a_{\text{avg}} = \frac{\Delta v}{\Delta t}\]
Where \(\Delta v\) is the change in velocity and \(\Delta t\) is the time taken for this change.
- From 2 minutes to 8 minutes, \(\Delta v = 2.20 \text{ m/s} - 0\), over 360 seconds, resulting in an average acceleration of approximately 0.0061 m/s².
- Likewise, for 3 minutes to 9 minutes interval, using the same change in velocity over the same time, the average acceleration remains approximately 0.0061 m/s².

Understanding how velocity changes help to comprehend the concept of acceleration, even when motion seems uniform.
Motion Graphs
Motion graphs allow us to visualize an object's movement over time. Analyzing graphs helps connect visual data to motion concepts like velocity and acceleration.

**Position vs. Time Graph**: This graph shows how far the man has moved at each moment in time. In this case: - A horizontal line at the beginning (0 to 300 seconds) represents standing still as there is no change in position. - A diagonal, upward line from 300 seconds onward illustrates constant motion with a steady increase in position. The slope represents average velocity.
**Velocity vs. Time Graph**: This graph reflects how the man's speed changes over time. - A flat line at zero for the first 300 seconds signifies no movement. - A flat line at 2.20 m/s from 300 seconds onwards shows constant velocity movement. With no change in line height here, the average acceleration is represented by small changes in speed when it first occurs.

By analyzing these graphs, you can easily see where changes in motion, such as standing still or moving, occur and how they relate to calculated values of average velocity and acceleration.

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Most popular questions from this chapter

The position function \(x(t)\) of a particle moving along an \(x\) axis is \(x=4.0-6.0 t^{2}\), with \(x\) in meters and \(t\) in seconds. (a) At what time and (b) where does the particle (momentarily) stop? At what (c) negative time and (d) positive time does the particle pass through the origin? (e) Graph \(x\) versus \(t\) for the range \(-5 \mathrm{~s}\) to \(+5 \mathrm{~s}\). (f) To shift the curve rightward on the graph, should we include the term \(+20 t\) or the term \(-20 t\) in \(x(t) ?(\mathrm{~g})\) Does that inclusion increase or decrease the value of \(x\) at which the particle momentarily stops?

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Suppose a rocket ship in deep space moves with constant acceleration equal to \(9.8 \mathrm{~m} / \mathrm{s}^{2}\), which gives the illusion of normal gravity during the flight. (a) If it starts from rest, how long will it take to acquire a speed one-tenth that of light, which travels at \(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s} ?(\mathrm{~b})\) How far will it travel in so doing?

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If the position of a particle is given by \(x=20 t-5 t^{3}\), where \(x\) is in meters and \(t\) is in seconds, when, if ever, is the particle's velocity zero? (b) When is its acceleration \(a\) zero? (c) For what time range (positive or negative) is \(a\) negative? (d) Positive? (e) Graph \(x(t), v(t)\), and \(a(t)\).

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