Tangent line to a particles path at a specific time?

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To find the orientation of the tangent line to a particle's path at t = 2.10 s, the position vector r, velocity v, and acceleration a were calculated, yielding r = -7.54 i + 70.8 j, v = 5.23 i - 148 j, and a = 12.6 i - 212 j. The angle of the velocity vector was initially miscalculated as -87 degrees, which was incorrect. The correct angle, determined by using the inverse tangent function and adjusting for the fourth quadrant, is 272 degrees from the positive x-axis in an anticlockwise direction. This adjustment involves adding 360 degrees to the negative angle to obtain the correct orientation. Understanding the quadrant in which the velocity vector lies is crucial for accurate angle determination.
badtwistoffate
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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?
 
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i got 272 degree from the + x-axis in anticlockwise direction.
 
how did u get that?
did you just do the inverse tan and adding to 360? does that make sense?
 
your v lies in the fourth quadrant.
 

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