How Does Acceleration Affect Constant Velocity in Particle Motion?

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The discussion focuses on the relationship between acceleration and constant velocity in particle motion, specifically in the xy plane. It establishes that the magnitude of velocity can remain constant only if the condition axvx + ayvy = 0 is satisfied, where ax and ay are the components of acceleration, and vx and vy are the components of velocity. The derivation involves taking the derivative of the velocity magnitude and applying the chain rule, leading to the conclusion that the acceleration must be orthogonal to the velocity vector for constant speed.

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A particle is moving in the xy plane with velocity v(t) = Vx(t)i + Vy(t)j
and acceleration a(t) = ax(t)i + ay(t)j
By taking the appropriate derivative, show that the magnitude of v can be constant only if axvx + ayvy = 0

I did it in a similar way to the other problem I had as it seems like a similar problem,
My thinking is:

h = (Vx(t)i)^2 + (Vy(t)j)^2

s(t) = sqrt(h)

dh/dt = d(vx^2 + vy^2) / dt
= d(vx)^2/dt + d(vy)^2/dt
= (d(vx^2)/dvx)(dvx/dt) + (d(vy^2)/dvy)(dy/dt)
= 2vx(dx/dt) + 2vy(dy/dt)

ds/dt = (ds/dh)(dh/dt)
= ((1/2)h^(-1/2))[(2vx(dvx/dt))+(2vy(dvy/dt))]

((1/2)h^(-1/2))[(2vx(dvx/dt))+(2vy(dvy/dt))] = 0

((1/2)h^(-1/2))[(2vx(ax))+(2vy(ax))] = 0

2((1/2)h^(-1/2))[vxax+vyax] = 0

axvx + ayvy = 0


Thanks
 
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