# simple acceleration question

• October 10th, 2007, 10:33 AM
Demen Tolden
simple acceleration question
Hi guys. I've got another simple physics question that I must understand wrong and I'm looking for some help to better explain it.

The acceleration of a moving object has a constant magnitude of a = 3.00 m/s^2. The object starts out from a dead stop at t = 0.00 s and moves in a straight-line path. How far will it have traveled from its starting point at t = 5.00 s?

I think to myself that this object will have a speed of 0 at t=0, 3 at t=1, 6 at t=2, 9 at t=3, 12 at t=4, 15 at t=5. To find the solution all that needs to be done is to add these six numbers to get 45 m, but apparently this is wrong. The correct answer is 37.5 m. Why is this?
• October 10th, 2007, 11:15 AM
Harold14370
Because it is not traveling that speed for the whole second, only at the end of that second. You need to look up the formula for distance as a function of acceleration and time. It should be in your textbook, or if not look it up on the internet.
• November 25th, 2007, 02:27 AM
fetus
Take the integral twice for the acceleration to get "1.5t^2"

And the answer is "1.5(5)^2 - 1.5(0)^2 = 37.5"
• November 25th, 2007, 02:31 AM
Demen Tolden
Yup, I have since picked up a calculus book and am happy to say that I am on my way to build a solid mathematical foundation.
• November 25th, 2007, 06:24 AM
Quantime
Quote:

Originally Posted by Demen Tolden
Yup, I have since picked up a calculus book and am happy to say that I am on my way to build a solid mathematical foundation.

So is that what fetus just did calculus? Because I don't get what it is.
• November 25th, 2007, 11:18 AM
Demen Tolden
As far as I know, calculus is about rates of change, or slopes of lines at a particular point. For example, without calculus we couldn't know the slope of y=x<sup>2</sup> when x = 2. With a linear equation this is a simple matter since the entire line has the same slope, or in other words the rate of change never varies.

To find the slope of a linear equation at a particular point, we just take two points on the line and divide the change of the y value by the change of the x value, or in other words:

m = ( y<sub>1</sub> - y<sub>2</sub> ) / ( x<sub>1</sub> - x<sub>2</sub>)

where m is the slope of the line. However to get an accurate slope of a parabola, or other such line where the slope varies, we need to take increasingly small measurements of x. In other words if we want to find the slope of the equation y=x<sup>2</sup> when x = 2 we could take the slope of a line connecting the points

(2, 4) and (4,16) and get m = 6

and continue taking slopes as x gets increasingly smaller

with (2, 4) and (3, 9), m = 5
with (2, 4) and (2.5, 6.25), m = 4.5
with (2, 4) and (2.25, 5.0625), m = 4.25
with (2, 4) and (2.1, 4.41), m = 4.1
with (2, 4) and (2.01, 4.0401), m = 4.01

so we could probably see where this is going. The slope of the line y = x<sup>2</sup> approaches 4 as the distance between x<sub>1</sub> and x<sub>2</sub> approaches 0, where x<sub>1</sub> = 2. This is called a limit, and as far as I know is the main idea behind all calculus. I'd explain about the relation between this acceleration equation and calculus, but I'm going to be late for work.
• November 25th, 2007, 11:38 AM
Quantime
Quote:

Originally Posted by Demen Tolden
As far as I know, calculus is about rates of change, or slopes of lines at a particular point. For example, without calculus we couldn't know the slope of y=x<sup>2</sup> when x = 2. With a linear equation this is a simple matter since the entire line has the same slope, or in other words the rate of change never varies.

To find the slope of a linear equation at a particular point, we just take two points on the line and divide the change of the y value by the change of the x value, or in other words:

m = ( y<sub>1</sub> - y<sub>2</sub> ) / ( x<sub>1</sub> - x<sub>2</sub>)

where m is the slope of the line. However to get an accurate slope of a parabola, or other such line where the slope varies, we need to take increasingly small measurements of x. In other words if we want to find the slope of the equation y=x<sup>2</sup> when x = 2 we could take the slope of a line connecting the points

(2, 4) and (4,16) and get m = 6

and continue taking slopes as x gets increasingly smaller

with (2, 4) and (3, 9), m = 5
with (2, 4) and (2.5, 6.25), m = 4.5
with (2, 4) and (2.25, 5.0625), m = 4.25
with (2, 4) and (2.1, 4.41), m = 4.1
with (2, 4) and (2.01, 4.0401), m = 4.01

so we could probably see where this is going. The slope of the line y = x<sup>2</sup> approaches 4 as the distance between x<sub>1</sub> and x<sub>2</sub> approaches 0, where x<sub>1</sub> = 2. This is called a limit, and as far as I know is the main idea behind all calculus. I'd explain about the relation between this acceleration equation and calculus, but I'm going to be late for work.

I'm guessing that you could find a function for this decreasing number, on a graph for instance. I must also say that I have come across what you have wrote before.
• November 25th, 2007, 12:51 PM
bit4bit
Quote:

As far as I know, calculus is about rates of change, or slopes of lines at a particular point.
That is only one half of caclulus, differentiation. The other half is 'integration' and the basic statement of calculus is that integration and differentiation are opposite processes.

What you need to look at are the 'equations of motion'. There are 4 or 5 basic equations involving initial velocity (u), final velocity (v), acceleration (a), displacement (s), and time (t). Of the top of my head I can only remember two:

v^2 = u^2 + 2as

s = ut + (1/2)at^2

There are a couple more too. They are all derived using calculus, something which you should probably explore on your own...it will get you much more comfortable with understanding basic calculus and its applications.

For your application, look at what variables you are provided with, and which variable you need to obtain:

Quote:

The acceleration of a moving object has a constant magnitude of a = 3.00 m/s^2. The object starts out from a dead stop at t = 0.00 s and moves in a straight-line path. How far will it have traveled from its starting point at t = 5.00 s?
u = 0 m/s
a = 3 m/s^2
t = 5s (time interval)
s = ?

s = ut + (1/2)at^2
= 0*5 + (1/2)*3*25
= (1/2)*3*25
= (1/2)*75
= 37.5m

That is using the equation of motion stated above. For the calculus method, look at velocity-time graphs ( in this case the graph will be a straight line, because the gradient, v/t = a, is constant). Considering that integration will give you the area under the graph, you will arrive at a statement for v*t giving you the expression above for displacement. You will also find that it's not as simple as s=v*t, because it is specific to your graph - which will be a triangle, and thats where the (1/2) comes into it. As I said though, its better to derive the formulas yourself to get a better understanding.
• November 25th, 2007, 01:00 PM
Quantime
Quote:

Originally Posted by bit4bit
Quote:

As far as I know, calculus is about rates of change, or slopes of lines at a particular point.
That is only one half of caclulus, differentiation. The other half is 'integration' and the basic statement of calculus is that integration and differentiation are opposite processes.

What you need to look at are the 'equations of motion'. There are 4 or 5 basic equations involving initial velocity (u), final velocity (v), acceleration (a), displacement (s), and time (t). Of the top of my head I can only remember two:

v^2 = u^2 + 2as

s = ut + (1/2)at^2

There are a couple more too. They are all derived using calculus, something which you should probably explore on your own...it will get you much more comfortable with understanding basic calculus and its applications.

For your application, look at what variables you are provided with, and which variable you need to obtain:

Quote:

The acceleration of a moving object has a constant magnitude of a = 3.00 m/s^2. The object starts out from a dead stop at t = 0.00 s and moves in a straight-line path. How far will it have traveled from its starting point at t = 5.00 s?
u = 0 m/s
a = 3 m/s^2
t = 5s (time interval)
s = ?

s = ut + (1/2)at^2
= 0*5 + (1/2)*3*25
= (1/2)*3*25
= (1/2)*75
= 37.5m

That is using the equation of motion stated above. For the calculus method, look at velocity-time graphs ( in this case the graph will be a straight line, because the gradient, v/t = a, is constant). Considering that integration will give you the area under the graph, you will arrive at a statement for v*t giving you the expression above for displacement. You will also find that it's not as simple as s=v*t, because it is specific to your graph - which will be a triangle, and thats where the (1/2) comes into it. As I said though, its better to derive the formulas yourself to get a better understanding.

What does (1/2) mean? Half? Opposite to the square root?

MOD Edit: You don't have to quote the whole post to comment on such a small point, indeed I'd recommend you stop doing it like right now!
• November 25th, 2007, 01:06 PM
bit4bit
Quote:

Originally Posted by svwillmer
What does (1/2) mean? Half? Opposite to the square root?

Half,
1/2,
0.5
etc..

Put in brackets to avoid confusion...well that was the idea :)
• November 25th, 2007, 01:09 PM
Quantime
Quote:

Originally Posted by bit4bit
Quote:

Originally Posted by svwillmer
What does (1/2) mean? Half? Opposite to the square root?

Half,
1/2,
0.5
etc..

Put in brackets to avoid confusion...well that was the idea :)

:-D Thanks.
• December 23rd, 2007, 04:47 AM
olympic/science
What does (1/2) mean? Half? Opposite to the square root?

:idea: :) A BEAUTIFUL MIND