Math Problem Statement

What is the displacement in the x direction between 0 and 4 seconds as measured by a velocity-time graph?

Solution

To determine the displacement between 0 and 4 seconds from the given velocity-time graph, we need to find the area under the graph between those two time points. Displacement is the area under the velocity-time curve.

Step-by-step process:

  1. From 0 to 2 seconds:

    • The velocity starts at 12m/s-12 \, \text{m/s} and linearly increases to 6m/s-6 \, \text{m/s}.
    • The shape of the graph in this section forms a trapezoid.
    • The area of the trapezoid can be calculated as: Area1=12×(b1+b2)×h\text{Area}_1 = \frac{1}{2} \times (b_1 + b_2) \times h Where:
    • b1=12m/sb_1 = -12 \, \text{m/s} (initial velocity),
    • b2=6m/sb_2 = -6 \, \text{m/s} (final velocity at t=2t = 2 seconds),
    • h=2secondsh = 2 \, \text{seconds} (time duration). Area1=12×(12+(6))×2=12×(18)×2=18m.\text{Area}_1 = \frac{1}{2} \times (-12 + (-6)) \times 2 = \frac{1}{2} \times (-18) \times 2 = -18 \, \text{m}.
  2. From 2 to 4 seconds:

    • The velocity remains constant at 6m/s-6 \, \text{m/s}, forming a rectangle.
    • The area of the rectangle is: Area2=v×t=6m/s×2seconds=12m.\text{Area}_2 = v \times t = -6 \, \text{m/s} \times 2 \, \text{seconds} = -12 \, \text{m}.

Total Displacement:

Total Displacement=Area1+Area2=18m+(12m)=30m.\text{Total Displacement} = \text{Area}_1 + \text{Area}_2 = -18 \, \text{m} + (-12 \, \text{m}) = -30 \, \text{m}.

Thus, the displacement between 0 and 4 seconds is -30 meters, which corresponds to option E.

Would you like more details on this solution, or any clarifications?


Here are some related questions for further practice:

  1. How would the displacement change if the velocity remained constant at 12m/s-12 \, \text{m/s} for the entire 4 seconds?
  2. How do you calculate displacement from a velocity-time graph if the curve is non-linear?
  3. What does negative displacement indicate about the motion of the car?
  4. How can you find the acceleration from a velocity-time graph?
  5. What is the significance of the area under a velocity-time graph in physics?

Tip: Always pay attention to the shape of the graph (e.g., trapezoid, rectangle) when calculating areas for displacement.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Velocity-Time Graph
Displacement

Formulas

Area of a trapezoid: A = 1/2 * (b1 + b2) * h
Area of a rectangle: A = v * t
Displacement = Area under velocity-time graph

Theorems

Relationship between area under velocity-time graph and displacement

Suitable Grade Level

Grades 9-12