A particle is moving with one component of constant velocity parallel ...
When a body is moving parallel to the axis of y and moving parallel to x-axis then it's intersecting point is like a parabola .
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A particle is moving with one component of constant velocity parallel ...
Explanation:
Given that the particle has one component of velocity parallel to the y-axis (let's call it vy) that is constant and another component of velocity parallel to the x-axis (let's call it vx) that is proportional to y.
We can represent the velocity vector of the particle as V = vx i + vy j, where i and j are the unit vectors along the x and y-axes respectively.
Component of velocity along the x-axis:
Since vx is proportional to y, we can write vx = ky, where k is a constant of proportionality.
Component of velocity along the y-axis:
Given that vy is constant, it remains the same throughout the motion.
Velocity vector:
Substituting the values of vx and vy in the velocity vector equation, we have V = ky i + vy j.
Acceleration vector:
The acceleration of the particle is the derivative of the velocity vector with respect to time.
Since vy is constant, its derivative with respect to time is zero.
The derivative of ky with respect to time is d(ky)/dt = k(dy/dt) = k(vy) = kvx.
Therefore, the acceleration vector of the particle is a = kvx i.
Equation of motion:
Using the equation of motion, we know that the acceleration vector is the second derivative of the position vector with respect to time.
Let's assume the position vector of the particle as r = x i + y j, where x and y are the components of position along the x and y-axes respectively.
The acceleration vector can be written as a = d^2(r)/dt^2 = (d^2x/dt^2)i + (d^2y/dt^2)j.
Since the acceleration vector is kvx i, we can equate the components of the acceleration vector with kvx to get:
d^2x/dt^2 = kvx (1)
d^2y/dt^2 = 0 (2)
Solving the equations:
Integrating equation (1) twice with respect to time, we get:
x = kvt + c1 (3)
where c1 is the constant of integration.
Since vy is constant, integrating equation (2) twice with respect to time gives:
y = c2 + c3t (4)
where c2 and c3 are constants of integration.
Particle's path:
From equations (3) and (4), we can see that the path of the particle is a straight line in the x-y plane.
Therefore, the correct option is (a) Circle.