- #1
sagaradeath
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Homework Statement
The position r-> of a particle moving in an xy plane is given by r->=(2t^3-2t)i+(3-2t^4)j with r in meters and t in seconds. What is the angle between the positive direction of the x-axis and a line tangent to the particle's path at t = 3 s? Give your answer in the range of (-180 degrees ; 180 degrees ].
Homework Equations
The Attempt at a Solution
first, find the particle's position at t=3
plug t into the r equation, you should get r = (48)i + (-159)j
which translates to the point (48, -159)
now find the velocity vector at t=3 (we do this to get the line tangent to path of motion)
take the derivative of r with respect to t, and get dr/dt = (6t^2-2)i + (-8t^3)j
solve for velocity at t=3, which should be dr/dt = (52)i + (-216)j
rise over run to find the slope of the line, m=(-216)/(52) = -4.15
now use point slope form to get the equation of the tangent line in y = x form, using the point we solved for in the beginning
y= -216-52*x + 525/13
that line crosses the x-axis at the point (175/18,0)
now to find the angle. draw a triangle with a base on the x-axis from x= 175/18 to x=48, and going down from the x-axis to y = -216. to find the angle, do tan@=(-216)/(48 - 175/18)
@(i use this for theta)= -79.95083227 degrees
I just want to know if I'm doing this problem right. Can someone help me with the answer.