ExamBro
ExamBro
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 \(\ldots \ldots {m} / {s}^{2}\)

  1. A \(12\)
  2. B \(16\)
  3. C \(8\)
  4. D \(24\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(12\)

Step-by-step Solution

Detailed explanation

\({V}=\sqrt{5000+24 {x}}\) \(\frac{{d} {V}}{{dx}}=\frac{1}{2 \sqrt{5000+24 {x}}} \times 24=\frac{12}{\sqrt{5000+24 {x}}}\) \(\text { now } {a}={V} \frac{{dV}}{{dx}}\) \(\quad=\sqrt{5000+24 {x}} \times \frac{12}{\sqrt{5000+24 {x}}}\) \(\qquad {a}=12 {m} / {s}^{2}\)
Same subject
Explore more questions on app