 56
 6
 Homework Statement

A particle moves along a defined curve so that its acceleration tangential component: ##a_t=ks##, where k is a constant and s denotes the arc distance respect to a point Q.
a) Find an expression for the velocity as a function of s.
b) Supposing that at Q its velocity equals 3.6 m/s and at A (s=5.4 m) 1.8 m/s, find k and its curvature radius at A knowing that the acceleration has magnitude 3.0 m/s^2.
c) Find at which distance the particle inverts its movement.
 Homework Equations

##\vec v= v e_t##
##\vec a= a_t e_t + a_n e_n = \frac{dv}{dt} e_t + \frac {v^2}{\rho} a_n##
So I know that ##a_t = \frac{dv}{dt}=ks## and ##\frac{dv}{dt}=v\frac{dv}{ds}## then: $$v dv=ks ds \rightarrow (v(s))^2=ks^2+c$$ and using my initial conditions it follows that: $$(3.6)^2=c \approx 13$$ and $$(1.8)^2=135.4k \rightarrow k=1.8 \rightarrow (v(s))^2=131.8s$$
What bothers me is finding at which point it turns, I am having trouble even getting started, any help would be appreciated.
What bothers me is finding at which point it turns, I am having trouble even getting started, any help would be appreciated.