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2.Motion in Straight Line
hard
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$\frac{{{b^2}\tau }}{4}$
(JEE MAIN-2019)
Solution
$\begin{array}{l}
V = b\sqrt X \\
\frac{{dv}}{{dt}} = \frac{b}{{2\sqrt X }}\frac{{dx}}{{dt}}\,;\,a = \,\frac{{bv}}{{2\sqrt X }}\\
a = \frac{{b\left( {b\sqrt X } \right)}}{{2\sqrt X }}\,;\,\frac{{dv}}{{dt}} = a = \frac{{{b^2}}}{2}\,;\,v = \frac{{{b^2}}}{2}\tau
\end{array}$
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}$
Standard 11
Physics
