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3-2.Motion in Plane
hard
A particle starts from origin at $t=0$ with a velocity $5 \hat{i} \mathrm{~m} / \mathrm{s}$ and moves in $x-y$ plane under action of a force which produces a constant acceleration of $(3 \hat{i}+2 \hat{j}) \mathrm{m} / \mathrm{s}^2$. If the $x$-coordinate of the particle at that instant is $84 \mathrm{~m}$, then the speed of the particle at this time is $\sqrt{\alpha} \mathrm{m} / \mathrm{s}$. The value of $\alpha$ is___________.
A$673$
B$685$
C$756$
D$741$
(JEE MAIN-2024)
Solution
$ \mathrm{u}_{\mathrm{x}}=5 \mathrm{~m} / \mathrm{s} \quad \mathrm{a}_{\mathrm{x}}=3 \mathrm{~m} / \mathrm{s}^2 \quad \mathrm{x}=84 \mathrm{~m} $
$ \mathrm{v}_{\mathrm{x}}^2-\mathrm{u}_{\mathrm{x}}^2=2 \mathrm{ax} $
$ \mathrm{v}_{\mathrm{x}}^2-25=2(3)(84) $
$ \mathrm{V}_{\mathrm{x}}=23 \mathrm{~m} / \mathrm{s} $
$ \mathrm{v}_{\mathrm{x}}-\mathrm{u}_{\mathrm{x}}=\mathrm{a}_{\mathrm{x}} \mathrm{t} $
$ \mathrm{t}=\frac{23-5}{3}=6 \mathrm{~s} $
$ \mathrm{v}_{\mathrm{y}}=0+\mathrm{a}_{\mathrm{y}} \mathrm{t}=0+2 \times(6)=12 \mathrm{~m} / \mathrm{s} 4$
$ \mathrm{v}^2=\mathrm{v}_{\mathrm{x}}^2+\mathrm{v}_{\mathrm{y}}^2=23^2+12^2=673 $
$ \mathrm{v}=\sqrt{673} \mathrm{~m} / \mathrm{s}$
$ \mathrm{v}_{\mathrm{x}}^2-\mathrm{u}_{\mathrm{x}}^2=2 \mathrm{ax} $
$ \mathrm{v}_{\mathrm{x}}^2-25=2(3)(84) $
$ \mathrm{V}_{\mathrm{x}}=23 \mathrm{~m} / \mathrm{s} $
$ \mathrm{v}_{\mathrm{x}}-\mathrm{u}_{\mathrm{x}}=\mathrm{a}_{\mathrm{x}} \mathrm{t} $
$ \mathrm{t}=\frac{23-5}{3}=6 \mathrm{~s} $
$ \mathrm{v}_{\mathrm{y}}=0+\mathrm{a}_{\mathrm{y}} \mathrm{t}=0+2 \times(6)=12 \mathrm{~m} / \mathrm{s} 4$
$ \mathrm{v}^2=\mathrm{v}_{\mathrm{x}}^2+\mathrm{v}_{\mathrm{y}}^2=23^2+12^2=673 $
$ \mathrm{v}=\sqrt{673} \mathrm{~m} / \mathrm{s}$
Standard 11
Physics
