In summary, When an object is tossed, it experiences a constant negative net force due to the equation F=ma where acceleration is always negative. The convention for down being negative is used. For an object moving in the xy plane with equations x=-(5.00m)sin(ωt) and y=(4.00m)-(5.00m)cos(ωt), the components of velocity and acceleration can be found by taking the derivative and second derivative of the equations. The notation of seconds^-1 can be used when finding the derivative of x, which is represented by the omega symbol. The velocity vector is represented by \vec{v} = v_{x} \hat i + v_{y} \hat j
At t=0 sinusoidal starts at 0, and cosinusoidal starts at Amplitude. In this case 1 multiplied by the 5w.
#41
UrbanXrisis
1,196
1
Okay, I have one more problem. You're a great help, seriously. I can't get though physics without PF...anyways. There's a fish that is swimming in a horzontal plane. It has an initial velocity of vi=(4.00i+1.00j)m/s at a point in the ocean whose displacement from a certain rock is r=(10.0i-4.00j)m. After the fish swims with constant acceleration for 20s, it has a v=(20.0i-5.00j) m/s. The componet of acceleration would be...
x=(4.00i-20.0i)/20s
y=(1.00j-(-5.00j))/20s
something like this? acceleration is the change in velocity over change in time...am I heading in the right direction?
The componet for velocity is x’= -(5.00m) ω cos ωt and y’= (5.00m) ω sin ωt
At t=0 seconds, x’= -(5.00m) ω and y’=0m/s
The componet for acceleration is x’’=(5.00m)ω^2 sin ωt and y’’= (5.00m) ω^2 cos ωt
At t=0 seconds, x’’=0m/s^2 and y’’=(5.00m)ω^2
Well i graphed it in my calculator, i guess i was wrong, looks like a circle, i was thinking about it too, looked like an elipse expression for a polar coordinate, I'm not sure about the path. In the cartesian it forms an arc.
#54
UrbanXrisis
1,196
1
I'm trying to understand what you did, is this relating my first problem or my second problem?
I asked a friend (mathematician) he says it will form a cardiode in a polar, and half a circle in cartesian, that should be the path, i don't know how to explain it to you.. but i think the question asked maybe requires a simpler solution, and i must had misunderstood it.
#56
UrbanXrisis
1,196
1
wow, that is WAY beyond me. What question were you trying to answer?
The componet for velocity is x’= -(5.00m) ω cos ωt and y’= (5.00m) ω sin ωt
At t=0 seconds, x’= -(5.00m) ω and y’=0m/s
The componet for acceleration is x’’=(5.00m)ω^2 sin ωt and y’’= (5.00m) ω^2 cos ωt
At t=0 seconds, x’’=0m/s^2 and y’’=(5.00m)ω^2