JEE MAIN - Physics (2020 - 9th January Morning Slot - No. 4)
The distance x covered by a particle in one
dimensional motion varies with time t as
x2 = at2 + 2bt + c. If the acceleration of the particle depends on x as x–n, where n is an integer, the value of n is __________
x2 = at2 + 2bt + c. If the acceleration of the particle depends on x as x–n, where n is an integer, the value of n is __________
Answer
3
Explanation
x2 = at2 + 2bt + c .....(1)
$$ \Rightarrow $$ 2xv = 2at + 2b .....(2)
$$ \Rightarrow $$ v = $${{at + b} \over x}$$
differentiating (2) w.r.t. time
$$ \Rightarrow $$ xa' + v2 = a
Here a' is acceleration.
$$ \Rightarrow $$ a'x = a - v2 = a - $${\left[ {{{at + b} \over x}} \right]^2}$$
$$ \Rightarrow $$ a'x = $${{a{x^2} - {{\left( {at + b} \right)}^2}} \over {{x^2}}}$$
$$ \Rightarrow $$ a' = $${{a\left( {a{t^2} + 2bt + c} \right) - {{\left( {at + b} \right)}^2}} \over {{x^2}}}$$
$$ \Rightarrow $$ a' = $${{ac - {b^2}} \over {{x^3}}}$$
$$ \therefore $$ a' $$ \propto $$ $${1 \over {{x^3}}}$$ $$ \propto $$ x-3
$$ \therefore $$ n = 3
$$ \Rightarrow $$ 2xv = 2at + 2b .....(2)
$$ \Rightarrow $$ v = $${{at + b} \over x}$$
differentiating (2) w.r.t. time
$$ \Rightarrow $$ xa' + v2 = a
Here a' is acceleration.
$$ \Rightarrow $$ a'x = a - v2 = a - $${\left[ {{{at + b} \over x}} \right]^2}$$
$$ \Rightarrow $$ a'x = $${{a{x^2} - {{\left( {at + b} \right)}^2}} \over {{x^2}}}$$
$$ \Rightarrow $$ a' = $${{a\left( {a{t^2} + 2bt + c} \right) - {{\left( {at + b} \right)}^2}} \over {{x^2}}}$$
$$ \Rightarrow $$ a' = $${{ac - {b^2}} \over {{x^3}}}$$
$$ \therefore $$ a' $$ \propto $$ $${1 \over {{x^3}}}$$ $$ \propto $$ x-3
$$ \therefore $$ n = 3
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