(B) Given the equation of motion: $y = bx^2$. Differentiating with respect to time $t$: $\frac{dy}{dt} = 2bx \frac{dx}{dt}$. This implies $v_y = 2bx v

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(B) Given the equation of motion: $y = bx^2$. Differentiating with respect to time $t$: $\frac{dy}{dt} = 2bx \frac{dx}{dt}$. This implies $v_y = 2bx v_x$. Differentiating again with respect to $t$: $\frac{dv_y}{dt} = 2b \left( \frac{dx}{dt} \cdot v_x + x \cdot \frac{dv_x}{dt} \right)$. Since the acceleration in the $x$-direction is zero $(a_x = 0)$,we have $\frac{dv_x}{dt} = 0$. Thus,$a_y = 2b(v_x^2)$. Given $a_y = a$,we have $a = 2bv_x^2$. Solving for $v_x$: $v_x^2 = \frac{a}{2b}$,which gives $v_x = \sqrt{\frac{a}{2b}}$.

Class 11 Physics 3-2.Motion in Plane Mix Examples-Motion in Plane

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$A$ particle crossing the origin at time $t=0$,moves in the $xy$-plane with a constant acceleration $a$ in the $y$-direction. If the equation of motion of the particle is $y = bx^2$ (where $b$ is a constant),then its velocity component in the $x$-direction is

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A
$\sqrt{\frac{2b}{a}}$
B
$\sqrt{\frac{a}{2b}}$
C
$\sqrt{\frac{a}{b}}$
D
$\sqrt{\frac{b}{a}}$

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3-2.Motion in Plane

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Mix Examples-Motion in Plane

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$A$ particle crossing the origin at time $t=0$,moves in the $xy$-plane with a constant acceleration $a$ in the $y$-direction. If the equation of motion of the particle is $y = bx^2$ (where $b$ is a constant),then its velocity component in the $x$-direction is

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$A$ boy throws a cricket ball from the boundary to the wicket-keeper. If the frictional force due to air cannot be ignored,the forces acting on the ball at the position $X$ are represented by

$A$ particle has an initial velocity $(2\hat i + 3\hat j)$ and an acceleration $(0.3\hat i + 0.2\hat j)$. Its speed after $10\,s$ is:

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