A particle crossing the origin of co-ordinates at time $\mathrm{t}=0$, moves in the $x y$ -plane with a…

A particle crossing the origin of co-ordinates at time $\mathrm{t}=0$, moves in the $x y$ -plane with a constant acceleration a in the y-direction. If its equation of motion is $y=b x^{2}(b$ is a constant), its velocity component in the $\mathrm{x}$ -direction is
  1. $\sqrt{\frac{2 \mathrm{~b}}{\mathrm{a}}}$
  2. $\sqrt{\frac{a}{2 b}}$
  3. $\sqrt{\frac{a}{b}}$
  4. $\sqrt{\frac{b}{a}}$

Solution

$\mathrm{y}=\mathrm{bx}^{2}$
Differentiating w.r.t to $\mathrm{t}$ an both sides, we get $\frac{d y}{d x}=b 2 x \frac{d x}{d t}$
$\mathrm{v}_{\mathrm{y}}=2 \mathrm{bxv}_{\mathrm{x}}$
Again differentiating w.r.t to $t$ on both sides we get $\frac{\mathrm{dv}_{\mathrm{y}}}{\mathrm{dt}}=2 \mathrm{bv}_{\mathrm{x}} \frac{\mathrm{dx}}{\mathrm{dt}}+2 \mathrm{bx} \frac{\mathrm{dv}_{\mathrm{x}}}{\mathrm{dt}}=2 \mathrm{bv}_{\mathrm{x}}^{2}+0$
$\left[\frac{\mathrm{dv}_{\mathrm{x}}}{\mathrm{dt}}=0\right.$, because the particle has constant acceleration along y-direction] Now, $\frac{\mathrm{dv}_{\mathrm{y}}}{\mathrm{dt}}=\mathrm{a}=2 \mathrm{bv}_{\mathrm{x}}^{2}$
$v_{x}^{2}=\frac{a}{2 b}$
$v_{x}=\sqrt{\frac{a}{2 b}}$ ^

Asked in: JEE Mains - Motion In One Dimension - Test 2

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