Finding force magnitude, angle, and direction from position equations

[Jrlinton]

Homework Statement

A 0.25 kg particle moves in an xy plane according to x(t) = -15 + 2t - 4t3 and y(t) = 25 + 7t - 9t2, with x and y in meters and t in seconds. Find formulas at time t<35 sec for the (a) the magnitude and (b) the angle (relative to the positive direction of the x axis) of the net force on the particle (in radians), as well as (c) the angle of the particle's direction of travel (in radians)

Homework Equations

F=ma
Pythagorean thereom
Fx=Fcosθ
Fy=Fsinθ

The Attempt at a Solution

With the x(t) and y(t) being the decoupled position vector, I thought I would use Pythagorean Thereom to come up with the vector r
r=((-4t^3+2t-15)^2+(-9t^2+7t+25)^2)^0.5
r=(16t^6+65t^4-6t^3-397t^2+290t+850)^.5
approximating the square roots for the coeffecients:
r=4t^3+9.06t^2-2.45t^1.5-19.92t+17.03t^0.5+29.15
I have a strong suspicion that I went about this in a very wrong way

 

[Jrlinton]

As Force=mass*acceleration and the acceleration in each direction is the second derivative of the decoupled position function the acceleration functions would be ax(t)=-24t ay(t)=-18

 

[Jrlinton]

Making the force function=((576t^2+324)^.5)/4 ?

 

[ehild]

Making the force function=((576t^2+324)^.5)/4 ?

Yes.

 

[Jrlinton]

Okay so the angle of the force would then be .25 arctan(-18/-24t) ??

 

[Jrlinton]

or .25 arctan(3/4t) simplified and the angle of the velocity being arctan(-18t/.12t^2)=arctan(3/2t)

 

[Jrlinton]

Those last two were incorrect.

 

[gneill]

Pay attention to the quadrant that the acceleration vector lies in. The arctan function can be a trap when the signs of the x and y components are not both positive. You need to sort out the correct quadrant yourself and adjust the result accordingly.