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3-2.Motion in Plane
hard
A particle moves in $x-y$ plane with velocity $\vec v = a\widehat i\, + \,bx\widehat j$ where $a$ & $b$ are constants. Initially particle was at origin then trajectory equation is:-
A$y = \frac{a}{b}x - \frac{1}{2}b{x^2}$
B$y = x - \frac{{b{x^2}}}{{2a}}$
C$y = \frac{{b{x^2}}}{{2a}}$
DNone of above
Solution
$\mathrm{V}_{\mathrm{x}}=\mathrm{a} \quad \mathrm{V}_{\mathrm{y}}=\mathrm{bx}$
${\int_{0}^{x} \mathrm{d} \mathrm{x}=\int_{0}^{\mathrm{t}} \mathrm{dt}}$ ${\mathrm{V}_{\mathrm{y}}=\mathrm{b} \mathrm{at}}$
${\mathrm{x}=\mathrm{at}}$ ${\int_{0}^{\mathrm{y}}\mathrm{dy}=\int_{0}^{\mathrm{t}} \mathrm{td} \mathrm{t}}$
$y=a b \frac{(x / a)^{2}}{2}=\frac{b x^{2}}{2 a}$
${\int_{0}^{x} \mathrm{d} \mathrm{x}=\int_{0}^{\mathrm{t}} \mathrm{dt}}$ ${\mathrm{V}_{\mathrm{y}}=\mathrm{b} \mathrm{at}}$
${\mathrm{x}=\mathrm{at}}$ ${\int_{0}^{\mathrm{y}}\mathrm{dy}=\int_{0}^{\mathrm{t}} \mathrm{td} \mathrm{t}}$
$y=a b \frac{(x / a)^{2}}{2}=\frac{b x^{2}}{2 a}$
Standard 11
Physics
