JEE Mains · Physics · STD 11 - 2. motion in straight line
If the velocity of a body related to displacement \({x}\) is given by \(v=\sqrt{5000+24 {x}} \;{m} / {s}\), then the acceleration of the body is \(\ldots \ldots {m} / {s}^{2}\)
- A \(12\)
- B \(16\)
- C \(8\)
- D \(24\)
Answer & Solution
Correct Answer
(A) \(12\)
Step-by-step Solution
Detailed explanation
\({V}=\sqrt{5000+24 {x}}\) \(\frac{{d} {V}}{{dx}}=\frac{1}{2 \sqrt{5000+24 {x}}} \times 24=\frac{12}{\sqrt{5000+24 {x}}}\) \(\text { now } {a}={V} \frac{{dV}}{{dx}}\) \(\quad=\sqrt{5000+24 {x}} \times \frac{12}{\sqrt{5000+24 {x}}}\) \(\qquad {a}=12 {m} / {s}^{2}\)
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