Angle Between Accel & Velocity: Solving Circular Motion Problems

[negation]

Your intuition was correct. 2.4 + 3.3 = 5.7
The angle is zero.

check post#34, something could be wrong

 

[Chestermiller]

check post#34, something could be wrong

No way. It's definitely zero.

(2.4 + 3.3 cosθ)²+(3.3sinθ)²=5.7²

(5.76+ 10.89cos²θ+15.84cosθ)+10.89sin²θ)=32.49

5.76+10.89(cos²θ+sin²θ)+15.84cosθ=32.49

5.76+10.89 +15.84 cosθ = 32.49

15.84 cosθ = 15.84

cosθ = 1

θ = 0

 

[negation]

No way. It's definitely zero.

(2.4 + 3.3 cosθ)²+(3.3sinθ)²=5.7²

(5.76+ 10.89cos²θ+15.84cosθ)+10.89sin²θ)=32.49

5.76+10.89(cos²θ+sin²θ)+15.84cosθ=32.49

5.76+10.89 +15.84 cosθ = 32.49

15.84 cosθ = 15.84

cosθ = 1

θ = 0

I had an arithmetic error.
Θ = 0 is correct.
Thanks and I shall now review the question again and try to sketch a mental model so as to better cement my understanding.

 

[negation]

No way. It's definitely zero.

(2.4 + 3.3 cosθ)²+(3.3sinθ)²=5.7²

(5.76+ 10.89cos²θ+15.84cosθ)+10.89sin²θ)=32.49

5.76+10.89(cos²θ+sin²θ)+15.84cosθ=32.49

5.76+10.89 +15.84 cosθ = 32.49

15.84 cosθ = 15.84

cosθ = 1

θ = 0

Would I be right in sketching
On the x-axis Vx=3.3cos(theta) +2.4
And
On the y-axis Vy=3.3sin(theta)
With the hypothenus being 5.7?

 

[Chestermiller]

Would I be right in sketching
On the x-axis Vx=3.3cos(theta) +2.4
And
On the y-axis Vy=3.3sin(theta)
With the hypothenus being 5.7?

Yes, as long as it's considered schematic. But, in the actual case with θ=0, the triangle is going to be degenerate.

 

[negation]

The above is not done correctly. Sorry. θ is not a function of t, so sinθ and cosθ factor out of the integral.

If you integrate those equations, it goes like this:
$$v_x(3)-v_x(0)=\int_0^3{1.1\cosθdt}=1.1\cosθ\int_0^3{dt}=3.3\cosθ$$
or
$$v_x(3)=2.4+3.3\cosθ$$
Similarly,
$$v_y(3)=3.3\sinθ$$

Strange.
If one integrates 1.1cos(x) shouldn't one obtain 1.1sin(x)? The antiderivative of cos is sin.

 

[haruspex]

Strange.
If one integrates 1.1cos(x) shouldn't one obtain 1.1sin(x)?

Only if you integrate with respect to x. If you integrate with respect to something unrelated to x, x staying constant, then cos(x) is a constant. Why should it change to sin(x)?