# One dimensional motion magnitude

1. Aug 30, 2004

### SMS

An object is moving in a straight line with a constant acceleration. Its position is measured at three different times, as shown in the table below.
Time (s) | Position, (m)
56.40 | 9.700
58.20 | 19.042
60.00 | 38.428
Calculate the magnitude of the acceleration at t=58.20 s

I am not quite sure how to start this. I thought you could find the velocities between the three different measurements and use this formula a =(vf-vi)/t.

2. Aug 30, 2004

### needhelpperson

i had a question like this before. Since i didn't know calculus yet, our teacher just told us to draw the best that we could of a tangent line at t=58.2 to the curve you made with those 3 points.

Last edited: Aug 30, 2004
3. Aug 30, 2004

### PRodQuanta

Yep, easiest way is to input that data into a graphing calculator in the table function. Then, graph that line. After that, find that point on your graph and VIOLA!

Paden Roder

4. Aug 30, 2004

### SMS

Ok

OK, I'll put it in my Calc and see what I get.

Thanks PRodQuanta and needhelpperson.

5. Aug 30, 2004

### needhelpperson

oops, i forgot that a constant acceleration has a linear graph with velocity vs time. you can just figure out the change in velocity over the first 2 points and divide it over the change in time. sorry bout that...

6. Aug 30, 2004

### SMS

Yes!!!!!!!

Thanks needhelpperson that worked and really helped. I will remember that for next time.

7. Aug 31, 2004

### Tide

I think if you realize that $$x = x_0 + v_0 t - \frac{a t^2}{2}$$ then you can use your data to obtain a simple linear system of equations for the three unknowns $$x_0$$, $$v_0$$ and $$a$$. Then you can use your solution of this system to calculate the acceleration or whatever you want at any time!

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