car particle moves in xy plane with a velocity v is equal to 2 y ICAP + 4 j cap equation of the path followed by the particle is
you misread it. v is a vector
v = 2yi + 4j
v=2y^4
so what direction is the velocity in? It matters.
To determine the equation of the path followed by the particle, we need to integrate the given velocity vector. Integration will allow us to find the position vector of the particle as a function of time.
Given velocity vector v = 2y î + 4 ĵ, where î and ĵ are the unit vectors in the x and y directions, respectively.
Integrating the x-component of the velocity vector:
∫ dx/dt = ∫ 2y dt
⇒ x = 2yt + c1, where c1 is the constant of integration.
Integrating the y-component of the velocity vector:
∫ dy/dt = ∫ 4 dt
⇒ y = 4t + c2, where c2 is the constant of integration.
The equation of the path followed by the particle can be obtained by substituting the expression for y into the equation for x:
x = 2(4t + c2)t + c1
or,
x = 8t^2 + 2c2t + c1
Therefore, the equation of the path followed by the particle is x = 8t^2 + 2c2t + c1, where c1 and c2 are constants.
To find the equation of the path followed by the particle, we need to integrate the given velocity vector with respect to time.
The given velocity vector is v = 2y ̂ + 4j ̂, where ̂i and ̂j are the unit vectors in the x and y directions, respectively.
Integrating this velocity vector will give us the position vector.
Integrating with respect to time, we have:
∫v dt = ∫(2y ̂ + 4j ̂) dt
Integrating 2y with respect to t gives us 2yt + C1, where C1 is the constant of integration.
Integrating 4 with respect to t gives us 4t + C2, where C2 is another constant of integration.
Therefore, the position vector of the particle is given by r = (2yt + C1) ̂i + (4t + C2) ̂j.
In order to find the equation of the path, we need to eliminate the constants C1 and C2 by applying initial conditions or constraints given in the problem. Without any additional information, we cannot determine the specific equation of the path followed by the particle.