What Formula Calculates Stopping Distance with Negative Acceleration?

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Acceleration = -3m/s^2
Meters per second = 11.18

if i were to find the stopping distance what formula should i use?

Vo + at = Vf

d = Vo x t + 1/2 x at^2

Vf^2 = Vo^2 + 2ad
 
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missie said:
Acceleration = -3m/s^2
Meters per second = 11.18

if i were to find the stopping distance what formula should i use?

Vo + at = Vf

d = Vo x t + 1/2 x at^2

Vf^2 = Vo^2 + 2ad

The 11.18, is it the Vf or Vi (or both?). Don't use a formula with t because you are not given a time. I would use the last one, Vf^2-Vi^2=2ad since you have ALL that information.
 
the 11.18 is the vi. How can i do this problem can you show me step to step? i have like 10 of these and hope to see how to do this one first.
 
Simple rule:
What do you know = u, v and a (hint you know final velocity=0)
What don't you know = t
What are looking for = d
which one gives you 'd' in terms of u,v,a ?
 
using the 3rd formula i got..

vf^2 = 125 + 2ad

/2a both side

vf^2 / 2a = d

and then what else do i do?

and is this right?
 
Last edited:
You know about how to rearrange formulae?

vf2=vi2 + 2ad
2ad = vf2 - vi2 and since vf=0 ie stopped
d = - vi2 /2a

Hint, put the units in for each number and check they cancel to give metres
 
nvm this is what i got...

vf^2 = 2ad

127/2a = d

after that?

127a/2a

= 127 / 2

= 63.5?
 
how is vf 0 when it only tells me meters per second is 11.18 and acceleration -3m/s?
 
Well, the *stopping* distance is the distance over which the object travels before *stopping*. If it is stopped, then its final velocity is ZERO. That's what it means to come to a stop...to no longer have any speed, in any direction.

Edit: and just to complete the thought, vf is the symbol for final velocity in this problem.
 
  • #10
at the end i got d = -9 / 2a

what do i do after this?

-9a / 2a

= -9 / 2

d = -4.5

is that right?
 
  • #11
I would check your algebra if I were you. The square of the velocity is NOT equal to 9.
 
  • #12
d = 125 / 2 * -3

d = 125 / -6


d = -20.83

is this right?
 
  • #13
The numbers are right. The answer should not be negative because there are TWO negative signs, one on the top, and one on the bottom (the -3). So the two negative signs will cancel, if you did your algebra properly.
 
  • #14
Remember an answer in physics without units is WRONG.
In addition the units check that you got the formula right.
 
  • #15
mgb_phys i got -20.83. is that right?
 
  • #16
because it's like this

d= 11.18^2 / 2 * -3

= 125 / -6

= -20.83

how can it not be negative?
 
  • #17
d = vf2 - vi2 /2a

vi = 11.18 m/s
vi = 0 m/s
a = -3 m/s2

d = 0 - 11.182 m2s-2 /(2*-3) m/s2 = 20.8 m
 
  • #18
missie said:
mgb_phys i got -20.83. is that right?

Ummm...hello? Already answered this question!

cepheid said:
The numbers are right. The answer should not be negative because there are TWO negative signs, one on the top, and one on the bottom (the -3). So the two negative signs will cancel, if you did your algebra properly.
 
  • #19
missie said:
mgb_phys i got -20.83. is that right?

I think your original equation was wrong.

Vf2 = Vo2 - 2*a*d

The negative sign because your acceleration is negative (slowing)

This then rearranges to Vo2 = 2*a*d

No negative signs, and distance in the direction of the initial velocity.
 

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