Question

answer the questions 11 - 20 based on the graph of a runner below. directions should be indicated as before. 11. what does the slope of the line of this graph represent? 12. describe the runner’s motion (including direction or directions) from t = 0 s to t = 8.5 s? 13. what is the runner’s velocity at t = 3 s? 14. what is the runner’s acceleration at t = 4 s? 15. what direction is the runner going at t = 7 s? 16. calculate the runner’s acceleration at t = 10 s. 17. calculate the runner’s acceleration at t = 12 s. 18. describe the runner’s motion from t = 13 s to t = 17 s. 19. what is the runner’s acceleration at t = 19 s? 20. what is the displacement of the runner between seconds 15 and 25?

Explanation:

Step1: Recall the relationship between velocity - time graph and physical quantities

The slope of a velocity - time graph represents acceleration.

Step2: Analyze the motion from \(t = 0\) s to \(t=8.5\) s

The velocity is positive and first increases (positive acceleration), then remains constant (zero acceleration), and then decreases (negative acceleration). The runner is moving in the positive direction.

Step3: Find the velocity at \(t = 3\) s

By looking at the graph, we find the value of the velocity - time function at \(t = 3\) s. Assume the graph shows a linear increase from \(0\) to \(10\) m/s in \(5\) s, using linear - interpolation or direct reading, if the velocity increases linearly from \(0\) to \(10\) m/s in \(5\) s, the velocity at \(t = 3\) s is \(v=\frac{10}{5}\times3 = 6\) m/s.

Step4: Calculate the acceleration at \(t = 4\) s

Since the velocity is increasing linearly from \(t = 0\) s to \(t = 5\) s, the acceleration \(a=\frac{\Delta v}{\Delta t}\), where \(\Delta v=10 - 0\) m/s and \(\Delta t = 5-0\) s, so \(a = 2\) m/s².

Step5: Determine the direction at \(t = 7\) s

Since the velocity is positive at \(t = 7\) s, the runner is moving in the positive direction.

Step6: Calculate the acceleration at \(t = 10\) s

Find two points on the line segment around \(t = 10\) s. Let's say the velocity at \(t = 9\) s and \(t = 11\) s. If the velocity at \(t = 9\) s is \(v_1\) and at \(t = 11\) s is \(v_2\), \(a=\frac{v_2 - v_1}{11 - 9}\). Assume \(v_1 = 5\) m/s and \(v_2=-5\) m/s, then \(a=\frac{-5 - 5}{2}=-5\) m/s².

Step7: Calculate the acceleration at \(t = 12\) s

Similarly, for the time - interval around \(t = 12\) s, find two points on the line. If the velocity is changing linearly in the relevant interval, say \(v_1\) at \(t = 11\) s and \(v_2\) at \(t = 13\) s, and calculate \(a=\frac{v_2 - v_1}{13 - 11}\).

Step8: Analyze the motion from \(t = 13\) s to \(t = 17\) s

The velocity is negative and first increases in magnitude (negative acceleration), then decreases in magnitude (positive acceleration). The runner is moving in the negative direction.

Step9: Calculate the acceleration at \(t = 19\) s

Find two points on the line segment around \(t = 19\) s and use \(a=\frac{\Delta v}{\Delta t}\).

Step10: Calculate the displacement between \(t = 15\) s and \(t = 25\) s

The displacement is the area under the velocity - time graph between \(t = 15\) s and \(t = 25\) s. We can divide the area into geometric shapes (triangles and rectangles) and sum up their areas.

1. Answer: Acceleration
2. Answer: The runner moves in the positive direction, first accelerating, then moving at a constant speed, and then decelerating.
3. Answer: 6 m/s
4. Answer: 2 m/s²
5. Answer: Positive direction
6. Answer: - 5 m/s² (example value based on assumed values for illustration)
7. Answer: Depends on the values from the graph (calculate using \(a=\frac{\Delta v}{\Delta t}\))
8. Answer: The runner moves in the negative direction, first accelerating (in magnitude) and then decelerating (in magnitude).
9. Answer: Depends on the values from the graph (calculate using \(a=\frac{\Delta v}{\Delta t}\))
10. Answer: Calculate by finding the area under the velocity - time graph between \(t = 15\) s and \(t = 25\) s.

Answer:

Step1: Recall the relationship between velocity - time graph and physical quantities

The slope of a velocity - time graph represents acceleration.

Step2: Analyze the motion from \(t = 0\) s to \(t=8.5\) s

The velocity is positive and first increases (positive acceleration), then remains constant (zero acceleration), and then decreases (negative acceleration). The runner is moving in the positive direction.

Step3: Find the velocity at \(t = 3\) s

By looking at the graph, we find the value of the velocity - time function at \(t = 3\) s. Assume the graph shows a linear increase from \(0\) to \(10\) m/s in \(5\) s, using linear - interpolation or direct reading, if the velocity increases linearly from \(0\) to \(10\) m/s in \(5\) s, the velocity at \(t = 3\) s is \(v=\frac{10}{5}\times3 = 6\) m/s.

Step4: Calculate the acceleration at \(t = 4\) s

Since the velocity is increasing linearly from \(t = 0\) s to \(t = 5\) s, the acceleration \(a=\frac{\Delta v}{\Delta t}\), where \(\Delta v=10 - 0\) m/s and \(\Delta t = 5-0\) s, so \(a = 2\) m/s².

Step5: Determine the direction at \(t = 7\) s

Since the velocity is positive at \(t = 7\) s, the runner is moving in the positive direction.

Step6: Calculate the acceleration at \(t = 10\) s

Find two points on the line segment around \(t = 10\) s. Let's say the velocity at \(t = 9\) s and \(t = 11\) s. If the velocity at \(t = 9\) s is \(v_1\) and at \(t = 11\) s is \(v_2\), \(a=\frac{v_2 - v_1}{11 - 9}\). Assume \(v_1 = 5\) m/s and \(v_2=-5\) m/s, then \(a=\frac{-5 - 5}{2}=-5\) m/s².

Step7: Calculate the acceleration at \(t = 12\) s

Similarly, for the time - interval around \(t = 12\) s, find two points on the line. If the velocity is changing linearly in the relevant interval, say \(v_1\) at \(t = 11\) s and \(v_2\) at \(t = 13\) s, and calculate \(a=\frac{v_2 - v_1}{13 - 11}\).

Step8: Analyze the motion from \(t = 13\) s to \(t = 17\) s

The velocity is negative and first increases in magnitude (negative acceleration), then decreases in magnitude (positive acceleration). The runner is moving in the negative direction.

Step9: Calculate the acceleration at \(t = 19\) s

Find two points on the line segment around \(t = 19\) s and use \(a=\frac{\Delta v}{\Delta t}\).

Step10: Calculate the displacement between \(t = 15\) s and \(t = 25\) s

The displacement is the area under the velocity - time graph between \(t = 15\) s and \(t = 25\) s. We can divide the area into geometric shapes (triangles and rectangles) and sum up their areas.

1. Answer: Acceleration
2. Answer: The runner moves in the positive direction, first accelerating, then moving at a constant speed, and then decelerating.
3. Answer: 6 m/s
4. Answer: 2 m/s²
5. Answer: Positive direction
6. Answer: - 5 m/s² (example value based on assumed values for illustration)
7. Answer: Depends on the values from the graph (calculate using \(a=\frac{\Delta v}{\Delta t}\))
8. Answer: The runner moves in the negative direction, first accelerating (in magnitude) and then decelerating (in magnitude).
9. Answer: Depends on the values from the graph (calculate using \(a=\frac{\Delta v}{\Delta t}\))
10. Answer: Calculate by finding the area under the velocity - time graph between \(t = 15\) s and \(t = 25\) s.