JEE Mains · Physics · STD 11 - 2. motion in straight line
A particle is moving with speed \(v= b\sqrt x\) along positive \(x-\) axis. Calculate the speed of the particle at time \(t = \tau\) (assume that the particle is at origin at \(t = 0\) ).
- A \({b^2}\tau \)
- B \(\frac{{{b^2}\tau }}{2}\)
- C \(\frac{{{b^2}\tau }}{{\sqrt 2 }}\)
- D \(\frac{{{b^2}\tau }}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{{{b^2}\tau }}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l} V = b\sqrt X \\ \frac{{dv}}{{dt}} = \frac{b}{{2\sqrt X }}\frac{{dx}}{{dt}}\,;\,a = \,\frac{{bv}}{{2\sqrt X }}\\ a = \frac{{b\left( {b\sqrt X } \right)}}{{2\sqrt X }}\,;\,\frac{{dv}}{{dt}} = a = \frac{{{b^2}}}{2}\,;\,v = \frac{{{b^2}}}{2}\tau \end{array}\)
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