A particle is moving in the xy plane with velocity v(t) = Vx(t)i + Vy(t)j(adsbygoogle = window.adsbygoogle || []).push({});

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 similiar way to the other problem I had as it seems like a similiar 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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# Homework Help: Vector Kinematics Take 2

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