- #1
badtwistoffate
- 81
- 0
how do you find orientation of a line that is tangent to the particle's path at t = 2.10 s?(from the +x axis)
before this i found the position r of a particle moving in an xy plane is given by r = (1.00t3 - 8.00t) i + (7.00 - 4.00t4) j.
So I found r,v,a, for t=2.10s (all in m)
r=-7.54 i +-70.8 j
v=5.23 i + -148 j
a=12.6 i + -212 j
i first tried using v's i and j to make a triangle and find the angle from that by inverse tan (-148/5.23) and got -87degress or -212 but that was wrong...
any help?
before this i found the position r of a particle moving in an xy plane is given by r = (1.00t3 - 8.00t) i + (7.00 - 4.00t4) j.
So I found r,v,a, for t=2.10s (all in m)
r=-7.54 i +-70.8 j
v=5.23 i + -148 j
a=12.6 i + -212 j
i first tried using v's i and j to make a triangle and find the angle from that by inverse tan (-148/5.23) and got -87degress or -212 but that was wrong...
any help?