Math Problem Statement

The following graphic represents the marking of a speedometer of an automobile as a function of time. Sketch the corresponding graphs of the acceleration and the distance traveled by the car as a function of time. What is the average acceleration of the car between t = 0 and t = 1 min? What is the average acceleration of the car between t = 2 min and t = 3 min?

Solution

Let's analyze the provided graph and solve the questions.

Graph Interpretation

  • The graph represents velocity (v) in km/h as a function of time (t) in minutes.
  • Key points:
    • From t=0t = 0 to t=0.5t = 0.5 min, the velocity increases from 0 to 45 km/h.
    • From t=0.5t = 0.5 to t=2t = 2 min, the velocity remains constant at 45 km/h.
    • From t=2t = 2 to t=2.5t = 2.5 min, the velocity increases to 75 km/h.
    • From t=2.5t = 2.5 to t=4t = 4 min, the velocity remains constant at 75 km/h.
    • From t=4t = 4 to t=4.5t = 4.5 min, the velocity decreases to 0 km/h.

(a) Sketch of Acceleration and Distance Graphs

  1. Acceleration Graph:

    • Acceleration is the change in velocity over time.
    • From t=0t = 0 to t=0.5t = 0.5 min: The car accelerates from 0 to 45 km/h. The slope indicates positive acceleration.
    • From t=0.5t = 0.5 to t=2t = 2 min: No change in velocity, so acceleration is 0.
    • From t=2t = 2 to t=2.5t = 2.5 min: The velocity increases from 45 to 75 km/h, indicating positive acceleration.
    • From t=2.5t = 2.5 to t=4t = 4 min: No change in velocity, so acceleration is 0.
    • From t=4t = 4 to t=4.5t = 4.5 min: The car decelerates to 0 km/h, showing negative acceleration.
  2. Distance Graph:

    • Distance is the area under the velocity-time graph.
    • Calculate the area of each segment (rectangle or triangle) under the velocity curve to sketch the distance graph. Each segment represents the distance traveled over time.

(b) Average Acceleration between t=0t = 0 and t=1t = 1 min

  • Formula: aavg=ΔvΔta_{\text{avg}} = \frac{\Delta v}{\Delta t}
  • Initial velocity v0=0v_0 = 0 km/h, final velocity v=45v = 45 km/h, Δt=1\Delta t = 1 min = 160\frac{1}{60} h. aavg=4501/60=45×60=2700 km/h2a_{\text{avg}} = \frac{45 - 0}{1/60} = 45 \times 60 = 2700 \text{ km/h}^2

(c) Average Acceleration between t=2t = 2 and t=3t = 3 min

  • Initial velocity v0=45v_0 = 45 km/h, final velocity v=75v = 75 km/h, Δt=1\Delta t = 1 min = 160\frac{1}{60} h. aavg=75451/60=30×60=1800 km/h2a_{\text{avg}} = \frac{75 - 45}{1/60} = 30 \times 60 = 1800 \text{ km/h}^2

Would you like a detailed sketch of the acceleration and distance graphs? Let me know if you have any questions!

Related Questions:

  1. How would the graph of acceleration differ for a more gradual acceleration phase?
  2. How can the area under a velocity-time graph be used to find the exact distance traveled?
  3. How do you determine instantaneous acceleration from a velocity-time graph?
  4. What is the physical significance of negative acceleration on this graph?
  5. How would the velocity-time graph change if the car started decelerating from t=2t = 2 min instead of increasing speed?

Tip:

Always remember that the slope of a velocity-time graph represents acceleration, and the area under it represents distance traveled.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Acceleration
Velocity-Time Graph

Formulas

Average acceleration: a_avg = (Δv / Δt)
Distance traveled: Area under the velocity-time graph

Theorems

Kinematic Equations

Suitable Grade Level

College Level (Physics 101 or 107L)