## anonymous 5 years ago A particle moves under a hypothetical force so that its velocity vector is v= <ktcost, ktsint>. Find its radial acceleration...Please help...

1. anonymous

whata hard bout that?

2. anonymous

acceleration is the derivative of velocity with respect to time

3. anonymous

assuming k is a constant

4. anonymous

$a = < k \frac{d}{dt} ( tcos(t)) , k \frac{d}{dt} ( tsin(t) ) >$

5. anonymous

k can be factored out of the vector, and it will just provide a scalar multiple

6. anonymous

$a = k < ( \cos(t) -tsin(t)) , (\sin(t) + tcos(t)) >$

7. anonymous

by product rules

8. anonymous

wow you are such a genius. I admire you all

9. anonymous

?

10. anonymous

bt i need the radial accn, not the total accn....

11. anonymous

at time t=0

12. anonymous

yeh i dnt have clue on that :|

13. anonymous

radial is apparently equal to centripetal= v^2 / r

14. anonymous

but i dont have a clue

15. anonymous

hey i got it neway ..thnx

16. anonymous

$\vec{a}_n=<-kt\sin{t},kt\cos{t}>$