Question 69: Motion in 1D

Question

A particle is moving with speed v = b$$\sqrt x $$ along positive x-axis. Calculate the speed of the particle at
time t = $$	au $$(assume that the particle is at origin t = 0)

Accepted Answer

v = b$$\sqrt x $$
$$ \Rightarrow $$ $${{dx} \over {dt}}$$ = b$$\sqrt x $$
$$ \Rightarrow $$$$\int\limits_0^x {{{dx} \over {\sqrt x }}}  = \int\limits_0^t  {bdt} $$
$$ \Rightarrow $$$$\left[ {{{{x^{ - {1 \over 2} + 1}}} \over { - {1 \over 2} + 1}}} ight]_0^x$$ = $$b\left[ t ight]_0^t$$
$$ \Rightarrow $$ $${x^{{1 \over 2}}} = {{bt} \over 2}$$
$$ \Rightarrow $$ $$x = {{{b^2}{t^2}} \over 4}$$
$$ 	herefore $$ v =  $${{dx} \over {dt}}$$ = $${{{b^2}} \over 4}$$ $$ 	imes $$ 2t = $${{{b^2}t} \over 2}$$
When t = $$	au $$ then speed v  $$ = {{{b^2}	au } \over 2}$$

Answer

$${{{b^2}	au } \over {\sqrt 2 }}$$

Answer

$${{b^2}	au }$$

Answer

$${{{b^2}	au } \over 2}$$

Answer

$${{{b^2}	au } \over 4}$$

Motion in 1D

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)

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