How Does Acceleration Affect Constant Velocity in Particle Motion?

  • Thread starter Thread starter Destrio
  • Start date Start date
  • Tags Tags
    Kinematics Vector
AI Thread Summary
In particle motion within the xy plane, a particle's 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. This relationship is derived from the derivatives of velocity and position, indicating that any change in velocity must be balanced by the acceleration components. The discussion highlights the mathematical process of showing this condition through derivatives and simplifications. The conclusion emphasizes that maintaining a constant velocity requires specific relationships between acceleration and velocity components. Understanding this principle is crucial for analyzing motion in physics.
Destrio
Messages
211
Reaction score
0
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
 
Physics news on Phys.org
Looks good! Very nice!
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top