# Calculating time of an accerating object
**URL:** https://boards.straightdope.com/t/calculating-time-of-an-accerating-object/710204
**Category:** Factual Questions
**Created:** [January 18, 2015, 7:28pm UTC](https://boards.straightdope.com/t/calculating-time-of-an-accerating-object/710204 "2015-01-18T19:28:09Z")
**Posts on this page:** 12
**Page:** 1
### Author:  [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)
#### Post date: [January 18, 2015, 7:28pm UTC](https://boards.straightdope.com/t/calculating-time-of-an-accerating-object/710204/1 "2015-01-18T19:28:09Z")
Lets say an arrow started a 0 fps and after traveling 24" reached a speed of 200 fps. Would the time spent accelerating simply be based the average speed of the arrow which would be 100 fps?
---
### Author:  [@beowulff](https://boards.straightdope.com/u/beowulff)
#### Post date: [January 18, 2015, 7:44pm UTC](https://boards.straightdope.com/t/calculating-time-of-an-accerating-object/710204/2 "2015-01-18T19:44:24Z")
Impossible to know.
ETA: not enough information - the arrow could be accelerating the entire distance, or it could have had a sharp impulse it the beginning and have coasted the rest of the way, or anywhere in-between.
---
### Author:  [@Andy\_L](https://boards.straightdope.com/u/Andy_L)
#### Post date: [January 18, 2015, 7:46pm UTC](https://boards.straightdope.com/t/calculating-time-of-an-accerating-object/710204/3 "2015-01-18T19:46:07Z")
> [@HoneyBadgerDC](#):
>
> Lets say an arrow started a 0 fps and after traveling 24" reached a speed of 200 fps. Would the time spent accelerating simply be based the average speed of the arrow which would be 100 fps?
Yes, the average velocity (time average) for a period of constant acceleration is the average of the starting velocity (in this case, 0) and the final velocity.
---
### Author:  [@MonkeyMensch](https://boards.straightdope.com/u/MonkeyMensch)
#### Post date: [January 18, 2015, 7:47pm UTC](https://boards.straightdope.com/t/calculating-time-of-an-accerating-object/710204/4 "2015-01-18T19:47:27Z")
> [@HoneyBadgerDC](#):
>
> … Would the time spent accelerating simply be based the average speed of the arrow which would be 100 fps?
It certainly would, providing that the acceleration was constant.
---
### Author:  [@Andy\_L](https://boards.straightdope.com/u/Andy_L)
#### Post date: [January 18, 2015, 7:53pm UTC](https://boards.straightdope.com/t/calculating-time-of-an-accerating-object/710204/5 "2015-01-18T19:53:23Z")
> [@beowulff](#):
>
> Impossible to know.
>
> ETA: not enough information - the arrow could be accelerating the entire distance, or it could have had a sharp impulse it the beginning and have coasted the rest of the way, or anywhere in-between.
Good point - I assumed a constant acceleration.
---
### Author:  [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)
#### Post date: [January 18, 2015, 8:00pm UTC](https://boards.straightdope.com/t/calculating-time-of-an-accerating-object/710204/6 "2015-01-18T20:00:12Z")
I expect that the rate of acceleration increases because of the leverage changes the arrow has over the bending bow due to string angle. The change in the rate would likley be fairy uniform yet I dont know what that rate would be.
---
### Author:  [@Andy\_L](https://boards.straightdope.com/u/Andy_L)
#### Post date: [January 18, 2015, 8:53pm UTC](https://boards.straightdope.com/t/calculating-time-of-an-accerating-object/710204/7 "2015-01-18T20:53:54Z")
> [@HoneyBadgerDC](#):
>
> I expect that the rate of acceleration increases because of the leverage changes the arrow has over the bending bow due to string angle. The change in the rate would likley be fairy uniform yet I dont know what that rate would be.
If you assumed the rate of increase in acceleration was constant (a constant “jerk”), you could determine the average velocity from the information given, I think. If I have a chance tonight I’ll derive…
---
### Author:  [@Andy\_L](https://boards.straightdope.com/u/Andy_L)
#### Post date: [January 18, 2015, 9:15pm UTC](https://boards.straightdope.com/t/calculating-time-of-an-accerating-object/710204/8 "2015-01-18T21:15:59Z")
> [@Andy\_L](#):
>
> If you assumed the rate of increase in acceleration was constant (a constant “jerk”), you could determine the average velocity from the information given, I think. If I have a chance tonight I’ll derive…
Actually was pretty simple. If jerk (change in acceleration) is constant, and acceleration and velocity start at zero, then distance travelled is 1/6 j t[sup]3[/sup] and velocity is 1/2 t[sup]2[/sup] over the period the jerk, j, is applied. At the end of that period T, we know distance is 2 feet, and velocity is 200 fps, so we can calculate T as .03 seconds. Over that time period, the arrow moved 2 feet giving an average velocity of 66 2/3 fps.
---
### Author:  [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)
#### Post date: [January 18, 2015, 9:27pm UTC](https://boards.straightdope.com/t/calculating-time-of-an-accerating-object/710204/9 "2015-01-18T21:27:57Z")
> [@Andy\_L](#):
>
> Actually was pretty simple. If jerk (change in acceleration) is constant, and acceleration and velocity start at zero, then distance travelled is 1/6 j t[sup]3[/sup] and velocity is 1/2 t[sup]2[/sup] over the period the jerk, j, is applied. At the end of that period T, we know distance is 2 feet, and velocity is 200 fps, so we can calculate T as .03 seconds. Over that time period, the arrow moved 2 feet giving an average velocity of 66 2/3 fps.
This agrees with how I felt intuitively but was unsure.
---
### Author:  [@Chronos](https://boards.straightdope.com/u/Chronos)
#### Post date: [January 19, 2015, 12:01am UTC](https://boards.straightdope.com/t/calculating-time-of-an-accerating-object/710204/10 "2015-01-19T00:01:39Z")
I would actually expect the acceleration to start off large, and then decrease. When a bow is undrawn, it takes very little force to pluck the string a small amount, but when it’s fully drawn, it takes a very large force to hold it drawn. And the force should be proportional to the acceleration of the arrow.
---
### Author:  [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)
#### Post date: [January 19, 2015, 12:05am UTC](https://boards.straightdope.com/t/calculating-time-of-an-accerating-object/710204/11 "2015-01-19T00:05:34Z")
> [@Chronos](#):
>
> I would actually expect the acceleration to start off large, and then decrease. When a bow is undrawn, it takes very little force to pluck the string a small amount, but when it’s fully drawn, it takes a very large force to hold it drawn. And the force should be proportional to the acceleration of the arrow.
```
You might be right, the force is much larger but the leverage is lower and the power drops faster. There is also the momentum of the limbs which might cancel out.
```
A very well designed bow will take advantage of the lower string angles at the end of the stroke and slow down the outer limbs bleeding off the momentum. I haven’t really dealt much with the acceleration rate until very recently. I was always more concerned with what happened in the last few inches of the power stroke. You probably are right.
---
### Author:  [@Andy\_L](https://boards.straightdope.com/u/Andy_L)
#### Post date: [January 19, 2015, 12:21am UTC](https://boards.straightdope.com/t/calculating-time-of-an-accerating-object/710204/12 "2015-01-19T00:21:34Z")
> [@Chronos](#):
>
> I would actually expect the acceleration to start off large, and then decrease. When a bow is undrawn, it takes very little force to pluck the string a small amount, but when it’s fully drawn, it takes a very large force to hold it drawn. And the force should be proportional to the acceleration of the arrow.
So approximating this as starting with acceleration of a[sub]0[/sub] and a constant jerk sufficient to reduce acceleration to zero at the moment of release might be more appropriate than my previous approximation. Using that assumption gives an average velocity of 133.33… fps over the period of acceleration.