one dimensional motion

[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.

[needhelpperson]

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.

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.

[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

[SMS]

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

Thanks PRodQuanta and needhelpperson.

[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...

[SMS]

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

[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!