Derivative of position question

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SUMMARY

The position of a dragonfly flying parallel to the ground is described by the vector function r⃗ =[2.90m+(0.0900m/s²)t²]i^−(0.0150m/s³)t³j^. To determine when the velocity vector makes a 30.0° angle clockwise from the x-axis, the derivative of the position vector was calculated, yielding (0.18t)i^−(0.045t²)j^. By applying the tangent function, tan(30°) = 0.18t/0.048t², the solution for t was found to be approximately 69.28 seconds, although the accuracy of this result was questioned during the discussion.

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Homework Statement



Q:
The position of a dragonfly that is flying parallel to the ground is given as a function of time by r⃗ =[2.90m+(0.0900m/s2)t^2]i^− (0.0150m/s3)t^3j^ .

At what value of t does the velocity vector of the insect make an angle of 30.0 ∘ clockwise from the x-axis?
.

Homework Equations


none

The Attempt at a Solution


What I did so far: Well I'm pretty sure the first step is to take the derivative of the position vector. I did that and got (0.18t)i^2-(0.045t^2)j^2.

I tried to use tan30=0.18t/0.048t^2 and solve for t

I get t=69.28 but I'm not sure if that is 100% correct
 
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Check this part of the question:

At what value of t does the velocity vector of the insect make an angle of 30.0 ∘ clockwise from the x-axis?

What can you defer from this?

Also,
I believe your evaluation of "t" has some mistakes.
 
never mind I got it

was trying to do it too fast and ended up making a careless mistake

thanks
 
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