(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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Choose the correct alternative$(s)$.

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Two balls are thrown as shown in the figure and they reach the ground in the same time. What is the ratio of the maximum heights attained by them?

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At a given instant of time, the position vector of a particle moving in a circle with a velocity $\vec{v} = 3 \hat{i} - 4 \hat{j} + 5 \hat{k}$ is $\vec{r} = \hat{i} + 9 \hat{j} - 8 \hat{k}$. Its angular velocity $\vec{\omega}$ at that time is:

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Column-$I$ (Angle of projection) Column-$II$
$A. \theta = 45^{\circ}$ $1. \frac{K_h}{K_i} = \frac{1}{4}$
$B. \theta = 60^{\circ}$ $2. \frac{gT^2}{R} = 8$
$C. \theta = 30^{\circ}$ $3. \frac{R}{H} = 4\sqrt{3}$
$D. \theta = \tan^{-1}(4)$ $4. \frac{R}{H} = 4$
$K_i:$ initial kinetic energy,$K_h:$ kinetic energy at the highest point.

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$A$ particle is projected from the ground with an initial speed $v$ at an angle $\theta$ with the horizontal. The average velocity of the particle between its point of projection and the highest point of its trajectory is

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Column-$I$ (Angle of projection)	Column-$II$
$A. \theta = 45^{\circ}$	$1. \frac{K_h}{K_i} = \frac{1}{4}$
$B. \theta = 60^{\circ}$	$2. \frac{gT^2}{R} = 8$
$C. \theta = 30^{\circ}$	$3. \frac{R}{H} = 4\sqrt{3}$
$D. \theta = \tan^{-1}(4)$	$4. \frac{R}{H} = 4$

$A$ 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

Similar Questions

Choose the correct alternative$(s)$.

Two balls are thrown as shown in the figure and they reach the ground in the same time. What is the ratio of the maximum heights attained by them?

At a given instant of time, the position vector of a particle moving in a circle with a velocity $\vec{v} = 3 \hat{i} - 4 \hat{j} + 5 \hat{k}$ is $\vec{r} = \hat{i} + 9 \hat{j} - 8 \hat{k}$. Its angular velocity $\vec{\omega}$ at that time is:

$A$ particle is projected from the ground with an initial speed $v$ at an angle $\theta$ with the horizontal. The average velocity of the particle between its point of projection and the highest point of its trajectory is

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