Starting with the kinematics equation x = vot + ½ at^2, derive an expression 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 initial velocity vo in terms of x1, t2, x2, and t2. The derivation involves solving a system of two equations and two unknowns.

Question

Starting with the kinematics equation x = vot + ½ at^2, derive an expression 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 initial velocity vo in terms of x1, t2, x2, and t2. The derivation involves solving a system of two equations and two unknowns.

Answer 1

To derive an expression relating the acceleration and initial velocity of an object to two paired position and time measurements, we'll start by rearranging the kinematic equation x = vot + ½at² to solve for the initial velocity vo.

Step 1: Set up the first equation
Let's assume that an object's position at time t1 is x1. Then, we have:
x1 = vo(t1) + ½a(t1)² ----- Equation (1)

Step 2: Set up the second equation
Assume that the object's position at time t2 is x2. Then, we have:
x2 = vo(t2) + ½a(t2)² ----- Equation (2)

Step 3: Solve the system of equations
We will now solve the system of equations (1) and (2) to eliminate vo.

From Equation (1), we can rearrange it to solve for vo:
vo(t1) = x1 - ½a(t1)²

Substitute this expression for vo in Equation (2):
x2 = (x1 - ½a(t1)²)(t2) + ½a(t2)²

Expanding and simplifying this equation gives:
x2 = x1(t2) - ½a(t1)²(t2) + ½a(t2)²

Rearranging this equation to isolate the acceleration term gives:
x2 - x1(t2) = ½a(t2)² - ½a(t1)²

Using the difference of squares, we can simplify further. Recall that a² - b² = (a + b)(a - b).

(x2 - x1)(t2) = ½a(t2)² - ½a(t1)²
(x2 - x1)(t2) = ½a(t2 + t1)(t2 - t1)

Divide both sides of the equation by ½(t2 - t1):
a = (x2 - x1)(2)/(t2 - t1)(t2 + t1)

Therefore, we have derived the expression relating acceleration and the two paired position and time measurements:
a = (x2 - x1)(2)/(t2 - t1)(t2 + t1)

Note: It's important to mention that this derivation assumes a constant acceleration. If the acceleration is not constant, this expression may not be accurate.

Answer 2

To derive an expression relating the acceleration and initial velocity of an object to two paired position and time measurements, we will start with the kinematics equation:

x = vot + ½ at^2

Let's consider two positions and corresponding times: (x1, t1) and (x2, t2).

For the first position (x1, t1):

x1 = vo(t1) + ½ a(t1)^2 --------------- Equation 1

For the second position (x2, t2):

x2 = vo(t2) + ½ a(t2)^2 --------------- Equation 2

Our goal is to express the acceleration (a) and initial velocity (vo) in terms of x1, t1, x2, and t2.

To solve this system of two equations and two unknowns, we need to eliminate one of the variables (either a or vo) from the equations.

Let's eliminate the initial velocity (vo). Rearrange Equation 1 to solve for vo:

vo = (x1 - ½ a(t1)^2) / t1

Substitute this expression for vo in Equation 2:

x2 = [(x1 - ½ a(t1)^2) / t1](t2) + ½ a(t2)^2

Expand and simplify:

x2 = (x1t2 / t1) - ½ a(t1)(t2) + ½ a(t2)^2

Now, let's simplify further:

x2 = (x1t2 / t1) + ½ a(t2)^2 - ½ a(t1)(t2)

Multiply through by t1:

x2t1 = x1t2 + ½ a(t2)^2t1 - ½ a(t1)(t2)t1

x2t1 = x1t2 + ½ a(t2)^2t1 - ½ a(t1)^2(t2)

Rearrange this equation to isolate the acceleration term:

x2t1 - x1t2 = ½ a(t2)^2t1 - ½ a(t1)^2(t2)

2(x2t1 - x1t2) = a(t2)^2t1 - a(t1)^2(t2)

2(x2t1 - x1t2) = at1^2(t2) - at2^2(t1)

Finally, solve for the acceleration (a):

a = [2(x2t1 - x1t2)] / [t1^2(t2) - t2^2(t1)]

With this expression for a, you can substitute the given values of x1, t1, x2, and t2 to calculate the acceleration.