JEE MAIN - Physics (2019 - 12th April Evening Slot - No. 19)
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 t = 0)
$${{{b^2}\tau } \over {\sqrt 2 }}$$
$${{b^2}\tau }$$
$${{{b^2}\tau } \over 2}$$
$${{{b^2}\tau } \over 4}$$
Explanation
v = b$$\sqrt x $$
$$ \Rightarrow $$ $${{dx} \over {dt}}$$ = b$$\sqrt x $$
$$ \Rightarrow $$$$\int\limits_0^x {{{dx} \over {\sqrt x }}} = \int\limits_0^t {bdt} $$
$$ \Rightarrow $$$$\left[ {{{{x^{ - {1 \over 2} + 1}}} \over { - {1 \over 2} + 1}}} \right]_0^x$$ = $$b\left[ t \right]_0^t$$
$$ \Rightarrow $$ $${x^{{1 \over 2}}} = {{bt} \over 2}$$
$$ \Rightarrow $$ $$x = {{{b^2}{t^2}} \over 4}$$
$$ \therefore $$ v = $${{dx} \over {dt}}$$ = $${{{b^2}} \over 4}$$ $$ \times $$ 2t = $${{{b^2}t} \over 2}$$
When t = $$\tau $$ then speed v $$ = {{{b^2}\tau } \over 2}$$
$$ \Rightarrow $$ $${{dx} \over {dt}}$$ = b$$\sqrt x $$
$$ \Rightarrow $$$$\int\limits_0^x {{{dx} \over {\sqrt x }}} = \int\limits_0^t {bdt} $$
$$ \Rightarrow $$$$\left[ {{{{x^{ - {1 \over 2} + 1}}} \over { - {1 \over 2} + 1}}} \right]_0^x$$ = $$b\left[ t \right]_0^t$$
$$ \Rightarrow $$ $${x^{{1 \over 2}}} = {{bt} \over 2}$$
$$ \Rightarrow $$ $$x = {{{b^2}{t^2}} \over 4}$$
$$ \therefore $$ v = $${{dx} \over {dt}}$$ = $${{{b^2}} \over 4}$$ $$ \times $$ 2t = $${{{b^2}t} \over 2}$$
When t = $$\tau $$ then speed v $$ = {{{b^2}\tau } \over 2}$$
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