ExamBro
ExamBro
JEE Mains · Physics · STD 11 - Motion in straight line
  • JEE Mains 2021
  • Hard
  • Numerical

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

Pick an option to check your answer.
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