- #1
Phys_Boi
- 49
- 0
So if a particle is at (0,0) and has a velocity of <-3,0> and an acceleration of <-2,-4> then we would add them to get the new position...
With a time interval of one second (t=1) then due to velocity:
Δx = -3(1) = -3
Δy = 0(1) = 0
then due to acceleration:
Δx = ½ (-2) (1)^2 = ½ (-2) = -1
Δy = ½ (-4) (1)^2 = ½ (-4) = -2
so we can add these displacements to get a new position of (-4, -2)
So my question is: after we arrive at this point and the acceleration has affected the velocity (and assuming the acceleration continues) is there another velocity vector? Or is there just the acceleration? Basically, do you only use the velocity once and forget it? Or does it still exist, and if so, how do you calculate it?
Thank you for any help!
With a time interval of one second (t=1) then due to velocity:
Δx = -3(1) = -3
Δy = 0(1) = 0
then due to acceleration:
Δx = ½ (-2) (1)^2 = ½ (-2) = -1
Δy = ½ (-4) (1)^2 = ½ (-4) = -2
so we can add these displacements to get a new position of (-4, -2)
So my question is: after we arrive at this point and the acceleration has affected the velocity (and assuming the acceleration continues) is there another velocity vector? Or is there just the acceleration? Basically, do you only use the velocity once and forget it? Or does it still exist, and if so, how do you calculate it?
Thank you for any help!