Please explain your steps in extensive detail, including which formulas you use,

Question

Please explain your steps in extensive detail, including which formulas you use, and explain your calculations algebraically. I am stuck on the problem of finding velocity, so if you would rather not solve for acceleration that is perfectly fine. It won't affect my vote. I believe the solution should be for instantaneous velocity.

I am pretty lost, so if your instinct tells you that a step is intuitive, please assume that it is not and notate accordingly!

The position of an object moving with constant acceleration is measured each second for 11 consecutive seconds. Find the initial and final velocities (v0, v11) and the acceleration (a) of the object.
t (s) x (cm)
0 2.11
1 6.39
2 16.95
3 33.79
4 56.91
5 86.31
6 121.99
7 163.95
8 212.19
9 266.71
10 327.51
11 394.59

Explanation / Answer

With this kind of problem, having a constant acceleration means that the velocity increases by the same amount each second. For example, if the velocity was 5 at time 1, and 10 at time 2, then it would be 15 at time 3 (add 5 each second for every second).

Knowing this, you can use the following equation to setup a few cases:

d = Vi * t + (1/2) * a * t^2
Since t is always 1 seconds for each step, you get:

d = Vi + a/2

Also, you can use the equation:

Vf^2 = Vi^2 + 2 * a * d

For 0 to 1 interval, you get:

6.39 - 2.11 = V0 + a/2 for the d = vi + a/2 equation and
V1^2 = V0^2 + a * 2 * (6.39 - 2.11) for the other equation

These simplify down to:

V0 = 4.28 - a/2
V0^2 = V1^2 - 8.56a

If you substitute the value for V0 from the first equation into the second, you get:

(4.28 - a/2)^2 = V1^2 - 8.56a

Now, you can get your last equation using the points between t2 and t1:

d = Vi + a/2
16.95 - 6.39 = V1 + a/2

Solving this for V1, you get V1 = 10.56 - a/2

Now, you can substitute that into the equation from above to get:

(4.28 - a/2)^2 = (10.56 - a/2)^2 - 8.56a

Now, you can solve this equation for a:

-4.28a = 93.20 - 10.56a - 8.56a
a = 6.28

So acceleration is 6.28 cm/s^2

Now, to get V0, you can use the equation from above with the value for a:

V0 = 4.28 - a/2 = 4.28 - 6.28/2 = 1.14 cm/s

And to get V11, you can use:

Vf^2 = Vi^2 + 2*a*d
V11^2 = V0^2 + 2*a*d
V11^2 = 1.14^2 + 2 * 6.28 * (394.59 - 2.11)
V11 = 70.22 cm/s

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