JEE Mains · Physics · STD 11 - 2. motion in straight line
A particle is moving with constant acceleration \('a'.\) Following graph shows \(v^{2}\) versus \(x\) (displacement) plot. The acceleration of the particle is \(......{m} / {s}^{2}\)
- A \(100\)
- B \(20\)
- C \(14\)
- D \(1\)
Answer & Solution
Correct Answer
(D) \(1\)
Step-by-step Solution
Detailed explanation
\({y}={m} {x}+{C}\) \({v}^{2}=\frac{20}{10} {x}+20\) \({v}^{2}=2 {x}+20\) \(2 {v} \frac{{d} {v}}{{d} {x}}=2\) \(\therefore {a}={v} \frac{{d} {v}}{{d} {x}}=1\)
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