- #1

willworkforfood

- 54

- 0

Is it true that if the rate of decceleration is two times slower than the rate of acceleration for the same object, that it will need two times the distance to come to a stop as it did to accelerate?

Last edited:

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter willworkforfood
- Start date

- #1

willworkforfood

- 54

- 0

Last edited:

- #2

russ_watters

Mentor

- 21,937

- 8,973

The answer to this question depends on if this is a homework question...

- #3

willworkforfood

- 54

- 0

It is but it is for extra credit :rofl:

- #4

LeonhardEuler

Gold Member

- 860

- 1

- #5

Doc Al

Mentor

- 45,447

- 1,907

- #6

willworkforfood

- 54

- 0

Doc Al said:Figure it out for yourself! How does the distance required to accelerate from a speed of 0 to V (or decelerate from V to 0) depend on the acceleration?

I have no idea where to start or how the distance is related to the acceleration.

- #7

Doc Al

Mentor

- 45,447

- 1,907

- #8

willworkforfood

- 54

- 0

- #9

Doc Al

Mentor

- 45,447

- 1,907

That's a start: You have a relation connecting acceleration and velocity (and time). Now figure out how to get the distance. (Hint: Distance equals average speed X time.)willworkforfood said:I know that the equation for acceleration is final velocity - initial velocity divided by time

- #10

willworkforfood

- 54

- 0

The problem doesn't give me average speed or time though.

- #11

Doc Al

Mentor

- 45,447

- 1,907

You don't need actual values. You are trying to find thewillworkforfood said:The problem doesn't give me average speed or time though.

(1) a = v/t (the initial speed is zero)

I suggest you combine that with my "hint" about distance and average speed. (If something goes from 0 to V, what is its average speed?) Write my statement mathematically. Then you'll have two equations that you can combine to relate distance with acceleration.

- #12

willworkforfood

- 54

- 0

So equation 2 would be x = v*t and you substitute v = at and get

x = a*t^2

Is that right?

Wow I feel stupid thanks for the help on that.

x = a*t^2

Is that right?

Wow I feel stupid thanks for the help on that.

Last edited:

- #13

Doc Al

Mentor

- 45,447

- 1,907

No.... you need average speed, which is v/2, not v.willworkforfood said:So equation 2 would be x = v*t

When you combine the equations, don't eliminate v; instead, eliminate t. You want to end up with an equation relating v, a, and x.and you substitute v = at and get

x = a*t^2

- #14

willworkforfood

- 54

- 0

x=(vt)/2

Solving for t would be

t = 2x/v

Substituting that into equation 2 would be:

a= v * (v/2x) or a = v^2 / 2x

Meaning when acceleration is 2 times decceleration then you need half the distance to accelerate? Thanks I think I get it now, I am not worthy :P

- #15

Doc Al

Mentor

- 45,447

- 1,907

Note that this same equation can help you understand why, for a given deceleration, that if you are moving twice as fast it takes 4 times the distance to stop.

Share:

- Replies
- 6

- Views
- 683

- Last Post

- Replies
- 2

- Views
- 2K

- Replies
- 1

- Views
- 399

- Last Post

- Replies
- 11

- Views
- 1K

- Replies
- 0

- Views
- 2K

- Last Post

- Replies
- 15

- Views
- 2K

- Replies
- 2

- Views
- 404

- Last Post

- Replies
- 9

- Views
- 7K

- Replies
- 5

- Views
- 123

- Replies
- 2

- Views
- 621