Question

the position of a dragonfly that is flying parallel to the ground is given as a function of time by r=[2.90m+(0.0900m/s^2)t^2]i-(0.0150m/s^3)t^3j. (a) At what value of t does the velocity vector of the insect make an angle of 30.0 clockwise from the +x-axis? (b) At the time calculated in part (a), what are the magnitude and direction of the acceleration vector of the insect?

Answer 1

Position vector:
r(t) = (2.9 + 0.09t^2) i - 0.015t^3 j
Velocity vector is derivative of position:
v(t) = dr/dt = 0.18t i - 0.045t^2 j
a.) We're looking for the angle of the velocity vector, which we can find using our trig identity:
tan(theta) = opposite/adjacent
For our v(t) we get
tan(theta) = -0.045t^2 / 0.18t = -0.25t
theta(t) = arctan(-0.25t)
And we want to know at what time t does theta(t) = -30 degrees:
arctan(-0.25t) = -30
-0.25t = tan(-30)
t = tan(-30)/-0.025
t = 2.31s

b.) Acceleration vector is derivative of velocity vector:
a(t) = dv/dt = 0.18 i - 0.09t j
At t = 2.31s the acceleration vector is
0.18 i - (0.09)(2.31) j = 0.18 i - 0.208 j
In polar coordinates that's 0.275m/s^2 @-49.1 degrees.

Answer 2

@dmancine
thank you for your answer,but when I put the answer in masterphysics "-49.1 degrees" was + not negative.

Answer 3

I can't see how r(t), v(t), or a(t) would have a positive j component at t>0. Can you see where I might have gone wrong?

Answer 4

actually i didnt understand the part you found 49,1 degrees. how did you find it?

Answer 5

I converted the rectangular coordinate vector 0.18i - 0.208j into polar coordinates. Do you understand how to do that?