# Distance traveled by a specific acceleration

Discussion in 'Physics' started by Lightfire, Dec 11, 2012.

1. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Hello!

I am having doubt to this equation on how to get the distance traveled by a specific acceleration.

I know of equation $a=v_{i}t+\frac{1}{2}at^{2}$.

For example, I am falling a ball at 10 seconds, would that mean that the distance traveled is 490 meters?

$a=v_{i}t+\frac{1}{2}at^{2}$
=$a=(0)(0)+\frac{1}{2}(9.8 m/s)(10 s)^{2}$
=$a=\frac{1}{2}(9.8 m/s)(100 s)$
=$a=\frac{1}{2}980m$
=$a=490m$

I doubt so. Did I make a mistake?

Thanks so much!

2. ### mrmount Active Member

Dec 5, 2007
59
7
I am not sure what you mean by "falling the ball at 10 seconds", but what you have calculated is the distance a free falling ball would travel under gravity in 10 seconds (starting with a velocity of 0 m/s at time = 0 seconds).

3. ### WBahn Moderator

Mar 31, 2012
23,194
6,986
1) Thanks for using units. So few people do.

2) Learn to use them correctly.

You have the acceleration at 9.8m/s, but acceleration is in distance per time-squared (or velocity per second), so it should be 9.8m/(s^2). Then you square 10s and get 100s when it should be 100s^2. So you either made two units errors that just, coincidentally, cancelled themselves out (certainly possible), or you used sleight-of-hand to make them appear to work out (the more likely, in my experience).

The careful and consistent tracking of units is very possibly the single most power tool you have to avoid making errors and detecting them when you do.

3) You use 'a' as the variable on the left hand side (for the distance traveled) and you use 'a' on the right hand side (for the acceleration). Which is it?

4) Why do you doubt your answer? I certainly applaud you for asking if your answer makes sense. Again, so few people do. 500 meters seems like a long way -- and it is; it is something on the order of a 100 storie building, such as Teipei 101 in Taiwan. But now imagine going to the top of the building and chucking off a cannon ball (i.e., something dense to minimize the impact of air resistance) and now count off ten seconds. Does it really seem so amazing that it could reach the ground in that amount of time? If the disbelieve is coming from the math itself, consider how fast the object is traveling after 10 seconds -- namely 98m/s. So in just the final second it will have travelled something close to 100m, or 20% of the total distance.

One takaway is that, if you can sustain an acceleration for an extended period of time, great speeds are achievable. For instance, let's say that you are in a spacecraft that could sustain an acceleration of 0.1m/(s^2) but could do it for a full day. It would take you take you about 10s just to get up to about normal walking speed, but after an hour you would have gotten up to 360m/s and covered 648km. At the end of the full day, you would be at over 86km/s and have travelled the distance from here to the moon.

4. ### Papabravo Expert

Feb 24, 2006
11,745
2,483
The units for acceleration are not meters per second; they are meters per second per second. This can also be expressed as meters per second squared. The two terms on the right hand side both have units of length aka meters, since velocity (m/s) times time (s) gives meters and acceleration (m/s/s) times time squared (s^2) gives length. Like units can be added together.

5. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Sorry for that mistake and thanks for your correction.

$d=v_{i}t+\frac{1}{2}at^{2}$

wherein:

d = distance
$v_{i}$= initial velocity
t=time
a=acceleration

$(0)(0)+\frac{1}{2}(9.8m^{2})(10s)^{2}$

I have some questions

5^2=25, right? Not 25^2 because that equals to 625.

, from your message also.

6. ### WBahn Moderator

Mar 31, 2012
23,194
6,986
Yes, that's correct? What's your point?

If you have two variables, a and b, and they are multiplied together and then squared, you have

y = (ab)^2

This is the same as

y = (a^2)(b^2)

If a = 10 and b = second, then you have

y = (10^2)(second^2)

y = (100)(second^2)

The proper abbreviation for second is 's', so we should write this as

y = (100)(s^2)

Since exponentiation has higher precedence than multiplication,the parentheses are redundant, so we can just write it as

y = 100s^2

This does NOT mean the square of (100 seconds), but rather 100 (seconds squared).

When writing this by hand, it is usually very clear that the exponent only applies to the units because we write the 2 tightly bound to the unit designator.

I don't really have any idea what you were trying to say in this part of your response. If you are saying that you now understand why an answer of about 500m is not unreasonable, great. If you are saying that your original concern was, in fact, because it seemed like too high a distance to travel in 10 seconds, then thanks for the confirmation. If you meant something else, please elaborate.

7. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Then you square 10s and get 100s when it should be 100s^2

I am still wondering why it should be 100s^2? Aren't 10s^2 equals 100s, and not 100s^2. (I could be wrong, though)

Yes, I have doubt that it seemed too high distance to travel in 10 seconds.

So, what do you think would be the answer?

8. ### strantor AAC Fanatic!

Oct 3, 2010
5,010
3,029
10s * 10s is the same thing as 10 * 10 * s * s. Not 10 * 10 * s. This becomes more obvious when you look at it from a different angle. Consider that when you square something, you can take the square root of the product and get the same number.

$10^{2} = 100$, and $\sqrt{100} = 10$
$(10a)^{2} = 100a^{2}$, and $\sqrt{100a^{2}} = 10a$

if we did things your way, it would look like this:

We square 10a and get 100a
$(10a)^{2} = 100a$

Then we take the square root of 100a
$\sqrt{100a} = 10\sqrt{a}$

and we don't get the same thing we started with:
$10a$$10\sqrt{a}$

This is because:
$(10a)^{2} = (10^{2})(a^{2})$
and
$\sqrt{100a^{2}} = (\sqrt{100})(\sqrt{a^{2}})$
I think you already got the right answer but still don't trust yourself.

9. ### Markd77 Senior Member

Sep 7, 2009
2,803
596
This isn't a very rigorous explanation for starting at zero speed, but the speed at the end = a X t and the average speed is half of that = 1/2 X a X t
To get the distance just multiply average speed by time

10. ### WBahn Moderator

Mar 31, 2012
23,194
6,986
The thing you have to realize is that units are as much a part of a value as the numbers are.

Saying that your desk is 4 wide is meaningless. It is 4 feet wide. So the answer is made up of two parts, a magnitude and a unit, that are, in effect, multiplied together to yield 4ft.

If you have a piece of paper that is 3ft wide by 4ft long, then the area is

Area = 3ft * 4ft = 12ft^2

This can be broken out explicitly as

Area = 3ft * 4ft
Area = (3)(ft)(4)(ft)
Area = [(3)(4)][(ft)(ft)]
Area = (12)[(ft)^2]
Area = 12[ft^2]
Area = 12ft^2

The square ONLY applies to the unit because exponentiation has higher precedence than multiplication, just like multiplication has higher precedence that addition.

Using labels (i.e., units) as though they were factors is known as the "factor label" method and it is extraordinarily powerful once you get comfortable with it which, to be sure, takes a bit of time and practice.

11. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Sorry guys. Pardon me, I forgot about that. Sorry. So would the answer be 490 meters?

Edit: I forgot that $(10s)^{2}=10^{2}s^{2}$.

12. ### WBahn Moderator

Mar 31, 2012
23,194
6,986
No need to apologize. Making mistakes is how we learn best. Sad, but true.

Yes, the answer is 490m.

13. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
I can understand multiplication is higher than addition, but what actually do you mean with that sentence, especially the word precedence?

What's the difference between ft^2 than ft aside from implying ft^2 maybe a unit of area and ft is a unit of maybe of length?

Sorry for these kinds of questions.

Thanks a lot!

14. ### WBahn Moderator

Mar 31, 2012
23,194
6,986
Precedence means, basically, priority. If I have two operators, such as + and *, and * has higher precedence than +, it means that when I would otherwise have the choice of which to do first, I have to do the one with higher precedence first. If I want to do the lower precedence operation first, then I have to use parentheses to enforce that -- because parentheses have higher precedence than anything else.

Ft^2 normally indicates that two values having units of distance (feet, specifcally, in this case) were multiplied together. It is usually indicative the the resulting numerical value represents an area with those dimensions. But, in general, it is not always so cut and dried. For instance, ft-lbs (foot-pounds) can either be a unit of torque or a unit of energy. It just works out that way.

15. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Thanks.

This came to my mind: can 9.8m/s^2 be multiplied by 100s? Since s^2 and s are two different units, they can't be cancelled. So hypotethicaly, what's the product of these two?

16. ### WBahn Moderator

Mar 31, 2012
23,194
6,986
Can you multiply A/B^2 by B? Sure. This is, after all, just

{A/[(B)(B)]} * B

which becomes

[(A)(B)]/[(B)(B)]

which can be written as

[(A)/(B)]*[(B)/(B)]

Anything (except zero) divided by itself is 1, so

[(A)/(B)]*1

Removing the parentheses, which aren't needed, we have:

A/B

Bottom line, treat the labels as though there were any other factor in the equation and that they obey the same rules -- this is why it is called the factor-lable method.

Just as B/B^2 reduces to 1/B, so t00 does s/s^2 reduce to 1/s.

So (9.8m/s^2)*(100s) yields 980m/s

Since this is v=at, it is reassuring that an acceleration multiplied by a period of time yields a result that has the units of velocity.

Oct 5, 2010
690
21
18. ### WBahn Moderator

Mar 31, 2012
23,194
6,986
It's called Algebra.

The basic idea is that if you have two things that are equal, say

7*6 = 42

Then as long as you do the same thing to both sides, the equality between them still holds. So

(7*6) - 12 = 42 - 12 = 30

[(7*6) - 12]/6 = 30/6 = 5

{[(7*6) - 12]/6} + 5 = 5 + 5 = 10

So, if you evaluate {[(7*6) - 12]/6} + 5, you should get 10.

But this works equally well when we don't know what one of the values is and, instead, represent that unknown value with a variabe such as 'x':

For what value of 'x' is the following equation true?

{[(7*x) - 12]/6} + 5 = 10

{[(7*x) - 12]/6} + 5 - 5 = 10 - 5 = 5

{[(7*x) - 12]/6} * 6 = 5 * 6 = 30

(7*x) - 12 +12 = 30 + 12 = 42

(7*x)/7 = 42/7 = 6

x = 6