Question:
A particle moves such that its velocity-time graph is a straight line from \(0\ \text{m/s}\) at \(t=0\) to \(20\ \text{m/s}\) at \(t=10\ \text{s}\). The displacement in \(10\ \text{s}\) is:
A particle moves such that its velocity-time graph is a straight line from \(0\ \text{m/s}\) at \(t=0\) to \(20\ \text{m/s}\) at \(t=10\ \text{s}\). The displacement in \(10\ \text{s}\) is:
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In a velocity-time graph, displacement is always equal to the area under the graph.
Updated On: May 20, 2026
- \(50\ \text{m}\)
- \(100\ \text{m}\)
- \(200\ \text{m}\)
- \(150\ \text{m}\)
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The Correct Option is B
Solution and Explanation
Concept:
Displacement from a velocity-time graph is equal to the area under the velocity-time graph. Here, the velocity-time graph is a straight line from \(0\) to \(20\ \text{m/s}\), so the area is triangular.
Step 1: Identify base and height of the triangle.
Base of the triangle: \(\displaystyle 10\ \text{s}\) Height of the triangle: \(\displaystyle 20\ \text{m/s}\)
Step 2: Use area of triangle.
\(\displaystyle \text{Displacement}=\frac{1}{2}\times \text{base}\times \text{height}\) \(\displaystyle =\frac{1}{2}\times 10\times 20\) \(\displaystyle =100\ \text{m}\)
Step 3: Final conclusion.
Hence, the displacement in \(10\ \text{s}\) is: \(\displaystyle \boxed{100\ \text{m}}\)
Displacement from a velocity-time graph is equal to the area under the velocity-time graph. Here, the velocity-time graph is a straight line from \(0\) to \(20\ \text{m/s}\), so the area is triangular.
Step 1: Identify base and height of the triangle.
Base of the triangle: \(\displaystyle 10\ \text{s}\) Height of the triangle: \(\displaystyle 20\ \text{m/s}\)
Step 2: Use area of triangle.
\(\displaystyle \text{Displacement}=\frac{1}{2}\times \text{base}\times \text{height}\) \(\displaystyle =\frac{1}{2}\times 10\times 20\) \(\displaystyle =100\ \text{m}\)
Step 3: Final conclusion.
Hence, the displacement in \(10\ \text{s}\) is: \(\displaystyle \boxed{100\ \text{m}}\)
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