Determining the final velocity and acceleration magnitude traveling along an arc

[HRubss]

Homework Statement

The motorcycle is traveling at 40 m/s when it is at A. If the speed is then decreased at $$v'=-(0.05s)m/s^2$$, where s is in meters measured from A, determine its speed and acceleration when it reaches B. I attached a picture of the problem.

Relevant Equations

$$S = S_0 + v_0(t) + \frac{1}{2}at^2$$
$$v^2 = (v_0)^2 + 2a(\Delta S)$$
$$s = r\theta$$
$$a_n = \frac{v^2}{\rho}$$
$$a_t = v'$$

Problem Statement: The motorcycle is traveling at 40 m/s when it is at A. If the speed is then decreased at v'=-(0.05s)m/s^2, where s is in meters measured from A, determine its speed and acceleration when it reaches B. I attached a picture of the problem.
Relevant Equations: S = S_0 + v_0(t) + \frac{1}{2}at^2
v^2 = (v_0)^2 + 2a(\Delta S)
s = r\theta
a_n = \frac{v^2}{\rho}
a_t = v'

I figured since the motorcycle travels along an arc, I needed to get the arc length. s = 150m(60*\frac{\pi}{180}) = 157.08 .
Then since the tangential acceleration is constant, using the constant acceleration formula to find final velocity...
v = \sqrt{(40)^2+2(-0.05(157.08))(157.08)} but that gave me an imaginary number since the acceleration is negative? I'm not sure if this is the correct process. Any help is appreciated!

 

[HRubss]

Can this be moved to introduction physics homework help forum? I think its better suited for there, even though this from an Engineering Dynamics textbook.

EDIT: Thank you!

 

[haruspex]

since the tangential acceleration is constant,

It isn't, and your first two Relevant Equations aren't, since they only apply to constant acceleration.
I do not understand your calculation for "v". Although the label v is used it represents speed here, not velocity. What is the speed at A?

 

[HRubss]

It isn't, and your first two Relevant Equations aren't, since they only apply to constant acceleration.
I do not understand your calculation for "v". Although the label v is used it represents speed here, not velocity. What is the speed at A?

Oh! I see, because its a function of distance? So would ads = vdv be more appropriate?
My "v" came from the constant acceleration formula but since it isn't constant, this will not work.
The speed at A is 40 m/s?

EDIT:
Wait I figured it out!
Since acceleration isn't constant and we're given the acceleration as a function of time.
ads = vdv
Integrating both sides,
\int ads = \int vdv
which gives me the final velocity and to find the acceleration magnitude, I used
a = \sqrt{a_t^2 + a_n^2}

Thanks for the help!