# Math Help - Obtaining deceleration from speed, distance and time

1. ## Obtaining deceleration from speed, distance and time

Is it possible to obtain the rate of deceleration of an object, if you know it's original speed/sec, expected distance to travel, and expected time to take e.g.

Object A is travelling at 800 units per second, towards a point (Z) which is 400 units away. A is expected to reach this point in 1 second, as it is decelerating all the way (reaching 0 units/second upon reaching Z) at a constant rate.

I hope I've explained what I'm trying to do well enough.

2. Originally Posted by Flare
Is it possible to obtain the rate of deceleration of an object, if you know it's original speed/sec, expected distance to travel, and expected time to take e.g.
Why yes there is! Have you come across:
$a = (v - u)/t$?

When I do these I sometimes forget that I have v (0m/s) Because that is when it stops.

How I haven't been too confusing.

3. Are you ready for "Limits"?

Figure it out.

Without deceleration: 400/800 = 1/2 second We're not real interested in this result.

Try it half-way at a time.

200 / 800 + 200 / x = 1 ==> 1/2 + 200/x = 1 ==> x < 0. This one went badly, too, since we accidentally made the whole trip on the first piece.

Try it 1/4 of the way at a time.

100/800 + 100/(800-x) + 100/(800-2x) + 100/(800-3x) = 1

Now try it 1/8 of the way at a time.

1) Can you write the expression?
2) Can you solve it?

4. @Evales: Would you mind explaining what each part of that is?

I'm guessing
a = accel?
v = 0 (or initial speed)
u = ???
t = time?

@TKHunny: I haven't a clue about what you're saying

5. a = acceleration
v = is not the initial speed it is the final speed but since you are talking about deceleration it is 0 anyway.
u = initial speed
t = time

Sorry I forget of course the symbols aren't standard in every curriculum.

6. But it almost seems as if you are given too much information. As we don't use the distance.
Alternatively you could use

$a = (v^2 - u^2) / 2s$

Which would be using the distance traveled instead of the time. One second I'm going to check if they come up with the same answer.

(Edit: s = distance)

7. They answers they give are hugely different are you sure that it decelerates fully?
Are you sure that it takes 1 second?

8. Well, I'm doing this for a system I'm making for a Warcraft 3 mod.

The a = (v(squared) - u(squared))/2(distance) seems to work.

In the first test, at 600 speed, 400 distance and 1.5second time, I was only out .15 seconds on the time, and 9 units on the distance.

In the second test, at 600 speed, 400 distance and 1second time, I was out -18 units on the distance, and had the time perfect.

Those errors shouldn't make a big difference though (who's gonna notice 18 units when it's only about this _ long )

Thanks very much for the help

9. Wow Physics can be used for something fun? Who would have thought....

10. I know, it's pretty awesome

11. Originally Posted by Flare
Well, I'm doing this for a system I'm making for a Warcraft 3 mod.

The a = (v(squared) - u(squared))/2(distance) seems to work.

In the first test, at 600 speed, 400 distance and 1.5second time, I was only out .15 seconds on the time, and 9 units on the distance.

In the second test, at 600 speed, 400 distance and 1second time, I was out -18 units on the distance, and had the time perfect.

Those errors shouldn't make a big difference though (who's gonna notice 18 units when it's only about this _ long )

Thanks very much for the help
I am rather confused about all of this. Both equations you were given will give the same result if you put the same data into it. The only way that it will not is if you have inconsistent data. Since the $v^2 - u^2$ equation works, I suspect you have an incorrect time interval.

-Dan

12. Originally Posted by Flare
@TKHunny: I haven't a clue about what you're saying
It's an exploration!

For this one:

100/800 + 100/(800-x) + 100/(800-2x) + 100/(800-3x) = 1

I split the trip up into four pieces of equal length. The trip is 400, so this explains the 100 in each numerator.

We start out at 800 ups. This explains the 800s in the denominator.

You defined uniform decelleration, so I applied a deceleration of 'x' to each successive unit.

First 100, speed = 800 -- no deceleration
Second 100, speed = 800 - x -- one deceleration of unknown size
Third 100, speed = 800 - 2x -- one more deceleration of the same size.
Fourth 100, speed = 800 - 3x -- one more deceleration of the same size.

How long did the first piece take? 100/800 seconds
How long did the second piece take? 100/(800-x) seconds
How long did the third piece take? 100/(800-2x) seconds
How long did the fourth piece take? 100/(800-3x) seconds

The last requirement is that the entire trip takes 1 second.

100/800 + 100/(800-x) + 100/(800-2x) + 100/(800-3x) = 1

Finally, there is a nice solution, x = 195.087. Let's check it out.

First 100, speed = 800, time = 100/800 = 0.125
Second 100, speed = 800 - 195.087 = 604.913, time = 100/604.913 = 0.165
Third 100, speed = 800 - 2(195.087) = 409.826, time = 100/409.826 = 0.244
Fourth 100, speed = 800 - 3(195.087) = 214.739, time = 100/214.739 = 0.466

Checking: 0.125+0.165+0.244+0.466 = 1.000

Thus, a constant decelleration of 195.087 ups every 100 units produces a good approximation of 1.95087 per unit.

Now for 8s. There are 8 pieces of the trip, each measuring 50 units. Again, the first speed is 800 ups, the second 800-x, the third 800-2x, down to the last 800-7x. For this one, I get a constant decelleration of 87.317 ups every 50 units produces a good approximation of 87.317/50 = 1.74634 per unit.

16s? x = 41.621 and the approximation is 41.621/25 = 1.66484 per unit

Noting how things are going:

4 -- 1.95087
8 -- 1.74634
16 -- 1.66484

Is there a limit to this process? If we tried 1,000 pieces, what woud we get? Maybe 1,000,000? Does it just keep going down or will it level off?

In four steps, from 4 to 8, the decrease averaged 0.051133. In eight steps, from 8 to 16, the decrease averaged only 0.010187. It's decreasing more slowly. I'm beginning to suspect that there is a limit to how low it will go.

Note: (That is, if you managed to get this far.) Like I said, this is an exploration. I'm just playing around with the situation a bit and seeing what there is to see and pointing out what there is to point out. If you're making a Warcraft Mod, this suggests you may have sufficient time on your hands to be able to enjoy such an exploration. Try to remember that the learning of mathematics can be FUN!!

13. Here's the 5 formula that are useful in calculating linear motion:

$v = u + at$
$s = \frac12(u+v)t$
$s = ut + \frac12 at^2$
$s = vt - \frac12at^2$
$v^2 = u^2 + 2as$

Where:
v = Final Velocity
u = Initial (Start) Velocity
a = Acceleration
s = Displacement (Distance travelled in a given direction)
t = Time of motion

Combining the first and the second equation allowed the bottom three equations to be derived.