Starting with the kinematics equation x = vot + ½ at^2, derive an expresion relating the acceleration and initial velocity of an object to two paired position and time measurements. In other words, if an object's position is x1 at t1, find the acceleration a and the inital velocity vo in terms of x1, t2, x2, and t2. The derivation involves solving a system of two equations and two unknowns.

To derive an expression relating the acceleration and initial velocity of an object to two paired position and time measurements, we will use the kinematics equation x = vot + ½ at² and solve a system of equations.

Given:

Position at time t₁: x₁
Position at time t₂: x₂

We know that the position at any time t is given by the equation x = vot + ½ at². For the first position:

x₁ = vo(t₁) + ½ a(t₁)²

For the second position:

x₂ = vo(t₂) + ½ a(t₂)²

Now, we have two equations with two unknowns: vo (the initial velocity) and a (the acceleration). We can solve this system of equations to find the expressions for vo and a.

To start, let's rearrange the first equation to solve for vo:

vo(t₁) = x₁ - ½ a(t₁)²

Substitute this expression for vo in the second equation:

x₂ = (x₁ - ½ a(t₁)²)(t₂) + ½ a(t₂)²

Expanding and simplifying:

x₂ = x₁t₂ - ½ a(t₁)²t₂ + ½ a(t₂)²

Now, isolate the terms involving a:

x₂ - x₁t₂ = ½ a(t₂)² - ½ a(t₁)²t₂

Factor out a:

x₂ - x₁t₂ = ½ a[(t₂)² - (t₁)²t₂]

Now, divide both sides by ½[(t₂)² - (t₁)²t₂]:

(x₂ - x₁t₂) / ½[(t₂)² - (t₁)²t₂] = a

Simplifying further:

a = 2(x₂ - x₁t₂) / [(t₂)² - (t₁)²t₂]

Thus, we have derived the expression for acceleration, a, in terms of x₁, t₂, x₂, and t₂.

Please note that the initial velocity, vo, is not involved in this derived equation. To find vo, you would need an additional equation or information.