SATHEE: Kinematics Question 591

Kinematics Question 591

Question: A particle crossing the origin of co-ordinates at time t = 0, moves in the xy-plane with a constant acceleration a in the y-direction. If it’s equation of motion is $y = b x^{2}$ (b is a constant), it’s velocity component in the x-direction is

Options:

A) $\sqrt{\frac{2 b}{a}}$

B) $\sqrt{\frac{a}{2 b}}$

C) $\sqrt{\frac{a}{b}}$

D) $\sqrt{\frac{b}{a}}$

Show Answer

Answer:

Correct Answer: B

Solution:

[b] $y = b x^{2}$ . Differentiating w.r.t to t an both

$\frac{dy}{dx} = b2x \frac{dx}{dt} \Rightarrow v_{y} = 2 bx v_{x}$

Again differentiating w.r.t to t on both sides we get

$\frac{d v_{y}}{d t} = 2 b v_{x} \frac{d x}{d t + 2 b x} \frac{d v_{x}}{d t} = 2 b v_{x}^{2} + 0$

$\frac{d}{v_{x}} dt = 0,$

because the particle had constant acceleration along y-direction

$Now, \frac{d}{v_{y}} dt = {a=2bv}_{x^{2}};$

$v_{x^{2}} = \frac{a}{2 b}$

$\Rightarrow v_{x} = \sqrt{\frac{a}{2b}}$

Kinematics Question 591

Question: A particle crossing the origin of co-ordinates at time t = 0, moves in the xy-plane with a constant acceleration a in the y-direction. If it’s equation of motion is $y = b x^{2}$ (b is a constant), it’s velocity component in the x-direction is

Options:

Answer:

Solution:

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Kinematics Question 591

Question: A particle crossing the origin of co-ordinates at time t = 0, moves in the xy-plane with a constant acceleration a in the y-direction. If it’s equation of motion is y = bx2 (b is a constant), it’s velocity component in the x-direction is

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: A particle crossing the origin of co-ordinates at time t = 0, moves in the xy-plane with a constant acceleration a in the y-direction. If it’s equation of motion is $y = b x^{2}$ (b is a constant), it’s velocity component in the x-direction is