Math Problem Statement
Solution
To find the object's displacement at the 7.5-second mark from the velocity vs. time graph, we need to calculate the area under the velocity curve between 0 seconds and 7.5 seconds. Displacement is determined by integrating (or finding the area under) the velocity-time graph.
Let's analyze the graph:
- From 0 to 5 seconds, the velocity is decreasing linearly from 8 m/s to -4 m/s. This section forms a trapezoid.
- From 5 to 7.5 seconds, the velocity increases linearly from -4 m/s to 0 m/s, forming a triangle.
Step-by-step process:
-
From 0 to 5 seconds (Trapezoid): The area of a trapezoid is given by: where , , and .
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From 5 to 7.5 seconds (Triangle): The area of a triangle is given by: Here, the base is the time interval (from 5 to 7.5 seconds), and the height is the change in velocity, from -4 m/s to 0 m/s, which is 4 m/s.
Total Displacement:
Summing the areas of the two regions:
Thus, the displacement at 7.5 seconds is 15 meters.
Would you like more details or have any questions?
Here are some related questions:
- How is displacement different from distance traveled?
- What is the physical significance of the area under a velocity-time graph?
- How would the displacement change if the velocity were negative throughout the entire time?
- How can you calculate displacement when the velocity function is not linear?
- What would the total displacement be at the 15-second mark based on this graph?
Tip: The area under a velocity-time graph gives displacement, while the area above and below the time axis may cancel out, depending on the direction of velocity.
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Math Problem Analysis
Mathematical Concepts
Kinematics
Displacement
Velocity-Time Graphs
Formulas
Area of a trapezoid: (1/2) * (v1 + v2) * t
Area of a triangle: (1/2) * base * height
Theorems
Displacement is the area under a velocity-time graph.
Suitable Grade Level
Grades 9-12
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