Homework Help: 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!

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!