JEE MAIN - Physics (2020 - 3rd September Evening Slot - No. 20)
A particle is moving unidirectionally on a horizontal plane under the action of a constant power
supplying energy source. The displacement (s) - time (t) graph that describes the motion of the
particle is (graphs are drawn schematically and are not to scale) :
_3rd_September_Evening_Slot_en_20_1.png)
_3rd_September_Evening_Slot_en_20_2.png)
_3rd_September_Evening_Slot_en_20_3.png)
_3rd_September_Evening_Slot_en_20_4.png)
Explanation
$$P = Fv$$
$$P = m{{dv} \over {dt}}v$$
$$ \Rightarrow $$ $$vdv = {P \over m}dt$$
Integrating both sides, we get
$${V^2} = k't$$
$$ \Rightarrow $$ V = k"$$\sqrt t $$
$$ \Rightarrow $$ $${{ds} \over {dt}}$$ = k"$$\sqrt t $$
$$ \Rightarrow $$ $$\int {ds} = \int {k''\sqrt t } dt$$
$$ \Rightarrow $$ s $$ \propto $$ t3/2
$$P = m{{dv} \over {dt}}v$$
$$ \Rightarrow $$ $$vdv = {P \over m}dt$$
Integrating both sides, we get
$${V^2} = k't$$
$$ \Rightarrow $$ V = k"$$\sqrt t $$
$$ \Rightarrow $$ $${{ds} \over {dt}}$$ = k"$$\sqrt t $$
$$ \Rightarrow $$ $$\int {ds} = \int {k''\sqrt t } dt$$
$$ \Rightarrow $$ s $$ \propto $$ t3/2
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