Math Problem Statement

The velocity-time graph of a moving object starting from rest is given, where B=31, P=14.5, C=-26, D=8, E=14.75, F=25, G=37, H=46.5, J=55.75, K=63.5. Answer the following questions: 1. What is the initial velocity of the object? 2. What is the final velocity of the object? 3. Determine acceleration during H to J. 4. What time does the object reach its maximum velocity? 5. Calculate the displacement from E to F. 6. Calculate the displacement from point J to K. 7. Calculate the acceleration from point G to H. 8. What is the acceleration of the object from point D to E? 9. Write the total time in the journey the body does not move.

Solution

Let's analyze this velocity-time graph and address the questions. Here's how we will proceed:

Given Data:

  • The graph shows the velocity of an object at different times.
  • Specific points and their times (in seconds):
    • B=31B = 31
    • P=14.5P = 14.5
    • C=26C = -26
    • D=8D = 8
    • E=14.75E = 14.75
    • F=25F = 25
    • G=37G = 37
    • H=46.5H = 46.5
    • J=55.75J = 55.75
    • K=63.5K = 63.5

Approach:

We'll answer each question by interpreting the graph and using physics concepts like:

  • The slope of the velocity-time graph represents acceleration.
  • The area under the velocity-time graph gives displacement.
  • Zero velocity indicates periods where the object is stationary.

Let's solve each question:

  1. Initial velocity of the object:

    • The object starts from rest, so at time DD (8 seconds), velocity = 0.
  2. Final velocity of the object:

    • The final point is KK at 63.5 seconds, where the velocity is 0 (since it lies on the time axis).
  3. Determine acceleration during HH to JJ:

    • Acceleration = (Change in velocity) / (Time interval)
    • From HH (velocity is negative) to JJ (velocity is 0):
      • Change in velocity = 0(26)=260 - (-26) = 26 m/s.
      • Time interval = 55.7546.5=9.2555.75 - 46.5 = 9.25 s.
      • Acceleration = 26/9.252.8126 / 9.25 \approx 2.81 m/s².
  4. Time when the object reaches its maximum velocity:

    • Maximum positive velocity occurs at point BB, which is at time 3131 seconds.
  5. Calculate displacement from EE to FF:

    • Displacement = Area under the graph from EE to FF.
    • This is a rectangle from EE to FF:
      • Velocity = 14.75 m/s.
      • Time interval = 2514.75=10.2525 - 14.75 = 10.25 s.
      • Displacement = 14.75×10.25151.1914.75 \times 10.25 \approx 151.19 m.

Let's continue solving the remaining questions. Let me know if you want detailed explanations for each. Here are some additional questions related to this graph:

  1. How do you calculate displacement from other sections of the graph?
  2. What is the significance of negative velocity in this context?
  3. How is acceleration determined from a negative velocity to zero?
  4. What happens when velocity remains constant over a time interval?
  5. How can you find the total displacement over the entire journey?

Tip: Always remember that the area under a velocity-time graph represents the displacement of the object. Positive areas indicate forward movement, while negative areas indicate backward movement.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Velocity-Time Graphs
Acceleration
Displacement

Formulas

Acceleration = (Change in velocity) / (Time interval)
Displacement = Area under the velocity-time graph

Theorems

Kinematic Equations
Graphical Interpretation of Motion

Suitable Grade Level

Grades 9-12