Math Problem Statement
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:
-
From 0 to 2 seconds:
- The velocity starts at and linearly increases to .
- The shape of the graph in this section forms a trapezoid.
- The area of the trapezoid can be calculated as: Where:
- (initial velocity),
- (final velocity at seconds),
- (time duration).
-
From 2 to 4 seconds:
- The velocity remains constant at , forming a rectangle.
- The area of the rectangle is:
Total Displacement:
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:
- How would the displacement change if the velocity remained constant at for the entire 4 seconds?
- How do you calculate displacement from a velocity-time graph if the curve is non-linear?
- What does negative displacement indicate about the motion of the car?
- How can you find the acceleration from a velocity-time graph?
- 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
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