## anonymous 4 years ago If a car travelling at vm/s takes X meters to stop, what would the stopping distance be if it was travelling at 3vm/s (assume stopping force is equal)

1. anonymous

I remember doing this in class, but forgot which equation to use

2. lgbasallote

hmm could it be $V_2 ^2 = V_1 + 2as$ where s is the distance

3. lgbasallote

correction.. $V_2 ^2 = V_1^2 + 2as$

4. anonymous

V is velocity? a is acceleration? s is displacement?

5. lgbasallote

yup

6. anonymous

Oh... yeah so that's the same one I have here as v^2=u^2+2as

7. lgbasallote

yep same shiz

8. lgbasallote

so first find the acceleration using v = 0 u = v s = X

9. anonymous

0=v^2+2aX... what is a?

10. lgbasallote

you solve for it

11. lgbasallote

isolate a

12. anonymous

a=-v^2/2X

13. lgbasallote

right

14. lgbasallote

now use v^2 = u^2 + 2as again this time use: v = 0 u = 3v a = -v^2/2X

15. lgbasallote

then solve for s

16. anonymous

starting with 0=9v^2+2aX -9v^2/2a=X -9v^2/2(-v^2/2X)=X 9X=X now what?

17. lgbasallote

no...

18. lgbasallote

it's not 2aX

19. lgbasallote

it's 2as

20. lgbasallote

X is just for the first situation

21. anonymous

oh i see

22. lgbasallote

$0 = 9v^2 + 2as$ $-9v^2 = 2as$ $-\frac{9v^2}{2a} = s$ $-\frac{9v^2}{2(-\frac{v^2}{2X})} = s$ $\implies -\frac{9v^2}{-\frac{v^2}{X}} = s$ $\implies 9X= s$

23. anonymous

Oh. THank You!

24. lgbasallote

welcome