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3.2 , motion in plane — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics3.2 , motion in plane1 Mark
MCQ
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
    $y = x - \frac{{b{x^2}}}{{2a}}$
  • ✓
    $y = \frac{{b{x^2}}}{{2a}}$
  • D
    None of above
✓

Answer

Correct option: C.
$y = \frac{{b{x^2}}}{{2a}}$
c
$\mathrm{V}_{\mathrm{x}}=\mathrm{a} \quad \mathrm{V}_{\mathrm{y}}=\mathrm{bx}$

${\int_{0}^{x} \mathrm{d} \mathrm{x}=\int_{0}^{\mathrm{t}} \mathrm{dt}}$               ${\mathrm{V}_{\mathrm{y}}=\mathrm{b} \mathrm{at}}$

${\mathrm{x}=\mathrm{at}}$          ${\int_{0}^{\mathrm{y}}\mathrm{dy}=\int_{0}^{\mathrm{t}} \mathrm{td} \mathrm{t}}$ 

 $y=a b \frac{(x / a)^{2}}{2}=\frac{b x^{2}}{2 a}$

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