# Get acceleration from a distance-time graph

• RuteNL
In summary, the conversation discusses a homework problem involving a graph with time on the x-axis and distance on the y-axis. The goal is to calculate acceleration using only the distance-time graph. The person attempted to use the equation S(t)=S(0) + V(0) * t + .5 * a * t2, but encountered discrepancies in their calculations. The conversation also touches on the accuracy of the graph and the possibility of using a polynomial fit to find the acceleration.
RuteNL

## Homework Statement

I got a graph with time on the x-axis, and distance on the y-axis. (see attached picture)
I have to calculate the acceleration using only the distance - time graph.

## Homework Equations

I used S(t)=S(0) + V(0) * t + .5 * a * t2

## The Attempt at a Solution

That gives 1.30=0+0*3.09 + .5 * a * 3.092
1.30/(.5/9.5481)=a
a=0.272 m/s2

But I know the acceleration is 0.42171 m/s2, because that's the slope of the V,t graph. What am I doing wrong here?

Thanks.

http://alecm.nl/user/ruurd/test/pfhelp.jpg

Last edited by a moderator:
The slope of the velocity time graph looks much less than 0.42. It looks to me (by eye) much more like 0.3.

Chet

The slope of the velocity I made roughly..

0.5/(3-1.8) = 0.42

so that seems ok.

I used S(t)=S(0) + V(0) * t + .5 * a * t2

That gives 1.30=0+0*3.09 + .5 * a * 3.092

The time axis doesn't appear to start at t=0 but at t=0.85.

At t=0.85 the position still appears to be approx 0 but the velocity appears to be 0.1m/s.

At t=2.8 the displacement appears to be 1m so..

1 = 0.1(2.8-0.85) + 0.5 a (2.8-0.85)^2

Gives a = 0.423

Chet [strike]is[/strike] was missing the fact that the x-axis doesn't start at zero.

I am impressed by the fancy teaching materials they have in NL. Now let's try to come up with a proper interpretation...

Can you indicate how you found s(0) and v(0) ?

Third time tonight quick replies cross. Starts to annoy me.
Isn't it bedtime for you Rute ? 0:30 AM ?!

Anyway, polynoom fit shows a3 = 0.21 which is 0.5 * 0.42, so data looks good.
Not only that, polynoom a2 is lineair a1, so that looks almost too good !

Don't know why the chi squareds are so low, but I suppose you don't have to worry about those in this stage of your physics education.

Back to the question how you found s(0) and v (0).

Since your mission is somewhat clearly formulated: "I have to calculate the acceleration using only the distance - time graph" I would be inclined to consider include the fit results as part of the Positie(Tijd) graph. But I am a lazy person. You Judge if that is allowed or not. If it is, life is easy: a1 = s(0), a2 = v(0), a3 = acceleration.
If it is not, there is some more work to do...

Did you really record this beautiful plot at 03:48 in the morning ? Wow...!

Last edited:
BvU said:
Chet [strike]is[/strike] was missing the fact that the x-axis doesn't start at zero.

You're right. Mia culpa.

Chet

## 1. What is acceleration?

Acceleration is the rate at which the velocity of an object changes over time. It is the change in velocity divided by the change in time.

## 2. How can acceleration be calculated from a distance-time graph?

Acceleration can be calculated from a distance-time graph by finding the slope of the line. The slope represents the change in distance over the change in time, which is equivalent to the change in velocity. Therefore, the slope is the acceleration.

## 3. What does a positive slope on a distance-time graph indicate?

A positive slope on a distance-time graph indicates that the object is moving with a constant acceleration in the positive direction. This means that the velocity of the object is increasing over time.

## 4. What does a negative slope on a distance-time graph indicate?

A negative slope on a distance-time graph indicates that the object is moving with a constant acceleration in the negative direction. This means that the velocity of the object is decreasing over time.

## 5. How does the steepness of the slope on a distance-time graph relate to acceleration?

The steepness of the slope on a distance-time graph directly relates to the magnitude of acceleration. A steeper slope indicates a greater change in velocity over time, and therefore a higher acceleration. A shallower slope indicates a smaller change in velocity over time, and therefore a lower acceleration.

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