Math Problem Statement

Example 88 A car travels at a constant velocity for 3 seconds. It then brakes consistently and comes to a halt in 2 seconds. Draw graphs of the car’s position, velocity, and acceleration vs. time.

Solution

To address this problem, we need to analyze the car's motion during the two phases: the constant velocity phase and the braking phase. We'll draw the position, velocity, and acceleration vs. time graphs for these phases.

Given:

  1. Constant velocity for 3 seconds.
  2. Braking phase for 2 seconds until the car comes to a halt.

1. Velocity vs. Time Graph

  • Phase 1 (0 to 3 seconds): The car moves at a constant velocity. The velocity graph will be a horizontal line.
  • Phase 2 (3 to 5 seconds): The car decelerates uniformly to zero. The velocity graph will be a straight line sloping downwards.

2. Acceleration vs. Time Graph

  • Phase 1 (0 to 3 seconds): Since the car is moving at constant velocity, the acceleration is 0. The acceleration graph will be a horizontal line on the time axis (acceleration = 0).
  • Phase 2 (3 to 5 seconds): The car decelerates uniformly. The acceleration will be negative and constant during this phase, represented by a horizontal line below the time axis.

3. Position vs. Time Graph

  • Phase 1 (0 to 3 seconds): The car is moving at constant velocity, so the position increases linearly with time. The graph will be a straight line with a positive slope.
  • Phase 2 (3 to 5 seconds): As the car decelerates, the slope of the position graph decreases until it reaches zero. The graph will curve, showing the deceleration until the position becomes constant when the car stops.

Now, I'll summarize how the graphs should look:

  1. Velocity vs. Time:

    • A horizontal line from t=0t = 0 to t=3t = 3 (constant velocity).
    • A straight line with negative slope from t=3t = 3 to t=5t = 5 (deceleration to zero).
  2. Acceleration vs. Time:

    • A horizontal line on the time axis from t=0t = 0 to t=3t = 3 (acceleration = 0).
    • A horizontal line below the time axis from t=3t = 3 to t=5t = 5 (constant negative acceleration).
  3. Position vs. Time:

    • A straight line with a positive slope from t=0t = 0 to t=3t = 3 (constant velocity).
    • A curve from t=3t = 3 to t=5t = 5 that flattens out as the car comes to a stop.

Would you like me to provide the graphs or need further details?


Relative Questions:

  1. How would the graphs change if the car accelerated instead of decelerating in the second phase?
  2. What would the position vs. time graph look like if the car started from rest?
  3. Can you describe the relationship between acceleration and velocity during the braking phase?
  4. How would you calculate the total distance traveled by the car?
  5. What is the significance of the area under the velocity vs. time graph?

Tip: Remember that the slope of the velocity vs. time graph represents acceleration, and the slope of the position vs. time graph represents velocity.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Graphical Analysis
Uniform Acceleration

Formulas

Velocity-time relationship for constant acceleration

Theorems

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Suitable Grade Level

Grades 11-12