Help me calculate the magnitude of the acceleration at t=57.10s?

Question

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)

56.10 9.300

57.10 12.150

58.10 18.300

Calculate the magnitude of the acceleration at t=57.10 s

I don't understand what type of acceleration it is asking for. I'm not sure if the third line of data is useful, or how to get acceleration with position and time and not velocity and time.

Answer 1

First get the velocity bbetween the two pairs of data, then the acceleration will follow:

v = delta(d)/ delta(t)

v_1 = (12.15-9.3)/(57.1-56.1) = 2.85/1 = 2.85 m/s

This is the average velocity from 56.1 to 57.1 seconds or "centered on" 56.6 seconds

v_2 = (18.3-12.15)/(58.1-57.1) = 6.15/1 = 6.15 m/s

This is the average velocity from 57.1 to 58.1 seconds or "centered on" 57.6 seconds

Now the acceleration is the change in velocity per unit time:

a = delta(v)/delta(t) = (6.15-2.85)/(57.6-56.6) = 3.3/1 = 3.3 m/s^2

This the average acceleration from 56.6 to 57.6 or "centered on" 57.1 seconds.

Hope this helps.

Answer 2

Graph it out. Pictures make things easy to visualize.

Acceleration is equal to m/(s^2)... I know it's hard to imagine seconds squared, but think about the formulas you have and not the fanciny numbers.

The third set of data is neccesary because you are looking for the change of acceleration between two points... Lemme see if i can illustrate.

Ready, set, GO!

Position1--- zoom zoom----Position2--- zoom zoom zoom----Position3

Yeah I know it's cheesey, but you're basically measuring the increase in "zooms" The guy before me explained it better, but this is the concept.