What Formula Calculates Stopping Distance with Negative Acceleration?

  • Thread starter Thread starter missie
  • Start date Start date
AI Thread Summary
To calculate stopping distance with negative acceleration, the appropriate formula is Vf^2 = Vi^2 + 2ad, where Vf is the final velocity (0 m/s when stopping), Vi is the initial velocity (11.18 m/s), and a is the acceleration (-3 m/s^2). By rearranging the formula, d can be expressed as d = (Vi^2) / (2 * -a), resulting in a positive distance value when calculated correctly. The discussion emphasizes that the final answer should not be negative, as two negative signs will cancel out. Proper unit checks are essential to ensure the calculations are accurate and meaningful in a physics context.
missie
Messages
38
Reaction score
0
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
 
Physics news on Phys.org
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.
 
Back
Top