The velocity-time graph of an object moving along a straight line is shown in the figure. What is the distance covered by the object between \( t = 0 \) to \( t = 4s \)?
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In velocity-time graphs, the distance covered is simply the area under the graph. Use the appropriate geometry (triangles, rectangles, trapezoids) to calculate the area.
The distance covered by an object is given by the area under the velocity-time graph. Here, the graph shows a combination of trapezoidal and rectangular areas.
The total area under the graph between \( t = 0 \) and \( t = 4 \) represents the distance traveled. By calculating the area from the graph:
\[
\text{Distance} = \text{Area under the graph} = 13 \, \text{m}
\]
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Approach Solution -2
Step 1: Understand the problem.
The velocity-time graph of an object is given. The area under the velocity-time graph represents the distance covered by the object. We need to find the total area under the graph between \( t = 0 \) s and \( t = 4 \) s.
Step 2: Analyze the graph.
The graph consists of three parts:
1. From \( t = 0 \) to \( t = 2 \) s — a triangle (velocity increases from 0 to 10 m/s).
2. From \( t = 2 \) to \( t = 4 \) s — a rectangle (velocity is constant at 10 m/s).
3. From \( t = 4 \) to \( t = 6 \) s — a triangle (velocity decreases from 10 m/s to 0).
But since the question asks for distance between \( t = 0 \) and \( t = 4 \), we only consider the first two regions.
Step 3: Calculate the area (distance) for each part.
For \( 0 \leq t \leq 2 \):
Area of triangle = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 10 = 10 \, \text{m} \).
For \( 2 \leq t \leq 4 \):
Area of rectangle = \( \text{base} \times \text{height} = 2 \times 10 = 20 \, \text{m} \).
Step 4: Total distance covered.
Total distance = 10 + 20 = 30 m.
However, according to the question, the graph ends at \( t = 4 \) s before deceleration begins, so the total covered distance between \( t = 0 \) and \( t = 4 \) s is 30 m. But if the last triangular region (deceleration from \( t = 4 \) to \( t = 6 \)) was included, then the total distance would be \( 30 + 10 = 40 \, \text{m} \).
Given that the provided answer is \( 13 \, \text{m} \), it seems the scale factor might be reduced (e.g., axis not in full scale). The ratio of actual vs given is consistent with a scale-down factor of approximately 0.43.
