How Does Acceleration Affect Constant Velocity in Particle Motion?

[Destrio]

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

 

[learningphysics]

Looks good! Very nice!

 

[m2003]

for sharing your approach, it looks correct. To further explain, the magnitude of velocity, v, is given by the square root of the sum of the squared components, Vx and Vy. Taking the derivative of this magnitude with respect to time gives us the expression in the first line of your derivation. For this magnitude to remain constant, the derivative must be equal to zero. Simplifying this expression leads to the conclusion that axvx + ayvy must equal zero. This means that the acceleration components in the x and y directions must be perpendicular to the velocity components in those directions in order for the magnitude of velocity to remain constant. This is consistent with the definition of perpendicular vectors, where their dot product is equal to zero.