JEE MAIN - Physics (2021 - 27th July Evening Shift - No. 15)
An automobile of mass 'm' accelerates starting from origin and initially at rest, while the engine supplies constant power P. The position is given as a function of time by :
$${\left( {{{9P} \over {8m}}} \right)^{{1 \over 2}}}{t^{{3 \over 2}}}$$
$${\left( {{{8P} \over {9m}}} \right)^{{1 \over 2}}}{t^{{2 \over 3}}}$$
$${\left( {{{9m} \over {8P}}} \right)^{{1 \over 2}}}{t^{{3 \over 2}}}$$
$${\left( {{{8P} \over {9m}}} \right)^{{1 \over 2}}}{t^{{3 \over 2}}}$$
Explanation
P = const.
$$P = Fv = {{m{v^2}dv} \over {dx}}$$
$$\int\limits_0^x {{P \over m}dx} = \int\limits_0^v {{v^2}dv} $$
$${{Px} \over m} = {{{v^3}} \over 3}$$
$${\left( {{{3Px} \over m}} \right)^{1/3}} = v = {{dx} \over {dt}}$$
$${\left( {{{3P} \over m}} \right)^{1/3}}\int\limits_0^t {dt} = \int\limits_0^x {{x^{ - 1/3}}} dx$$
$$ \Rightarrow x = {\left( {{{8P} \over {9m}}} \right)^{1/2}}{t^{3/2}}$$
$$P = Fv = {{m{v^2}dv} \over {dx}}$$
$$\int\limits_0^x {{P \over m}dx} = \int\limits_0^v {{v^2}dv} $$
$${{Px} \over m} = {{{v^3}} \over 3}$$
$${\left( {{{3Px} \over m}} \right)^{1/3}} = v = {{dx} \over {dt}}$$
$${\left( {{{3P} \over m}} \right)^{1/3}}\int\limits_0^t {dt} = \int\limits_0^x {{x^{ - 1/3}}} dx$$
$$ \Rightarrow x = {\left( {{{8P} \over {9m}}} \right)^{1/2}}{t^{3/2}}$$
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