Question

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)

Answer 1

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

Answer 2

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

Answer 3

correction.. \[V_2 ^2 = V_1^2 + 2as\]

Answer 4

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

Answer 5

yup

Answer 6

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

Answer 7

yep same shiz

Answer 8

so first find the acceleration using

v = 0
u = v
s = X

Answer 9

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

Answer 10

you solve for it

Answer 11

isolate a

Answer 12

a=-v^2/2X

Answer 13

right

Answer 14

now use

v^2 = u^2 + 2as again

this time use:
v = 0
u = 3v
a = -v^2/2X

Answer 15

then solve for s

Answer 16

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

Answer 17

no...

Answer 18

it's not 2aX

Answer 19

it's 2as

Answer 20

X is just for the first situation

Answer 21

oh i see

Answer 22

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

Answer 23

Oh. THank You!

Answer 24

welcome