1. Review the data collected in the lab and explain the movement of the cart relative to the reference point(starting point). Be sure to mention how far from the origin the cart gets in each time interval.

2. Review the data collected in the lab and determine the relationship between the position of the car relative to the reference point and the amount of time in each time interval in other words as time goes by how does the position change?

3. For this response he will show your work on calculating the speed of the car during the lab remember you are used to change and slope equation or (y^2-y^1)/(x^2-x^1).
Now let’s make connections to the data table the horizontal variable is the time which we will represent with t and the vertical variable is position which we will represent with an x using this we can rewrite the equation above to the following
(final position - initial position)/
(final time - initial time)

When the data of the position and time are available, the velocity which is the ratio of change in position to change in time which is the same as the slope of the graph of the data can be calculated. For the car, we have;

1. The car moves progressively further as the time interval increases

2. The relationship between the position of the car and the time interval is that the car moves approximately 0.02 units of distance for each change in time of 0.05

3. The speed of the car is 0.398

The reason for arriving at the above values are as follows:

1. From the given data, we have that the cart moves progressively further from the start for each time interval increase, such that the cart moves progressively further from the origin in each time interval increases

[tex]\begin{array}{|c|c|} \mathbf{ Time} & \mathbf{ Position}\\2.2&0.8963\\2.25& 0.9164\\2.3 & 0.9375\\2.35 & 0.9581\\2.4 & 0.9759\end{array}\right][/tex]

2. From the given data, we have;

At time t = 2.2, the position = 0.8963

At time t = 2.25, the position = 0.9164

The change in position for a change in time from 2.2 to 2.25 is, Δx = 0.9164 - 0.8963 = 0.0201

Similarly, for the change in time from 2.25 to 2.3, which is a change of 0.05 seconds, we have;

Δx = 0.9375 - 0.9164 = 0.0211

The next interval is 2.3 to 2.35, Δx = 0.9581 - 0.9375 = 0.0206

Therefore, the change in the distance from the origin increases by approximately 0.02 for each increase in tome by Δt = 0.05 seconds

3. The slope equation is presented as follows;

[tex]Slope = \dfrac{y_2 - y_1 }{x_2 - x_1}[/tex]

The speed of the car is given by the slope of the distance time graph, where;

The horizontal variable, x, is the time, t (x = t)

The vertical variable, y, is the position, x (y = x)

Based on the given assignment of the variables to the x, and y, variables, we have;

[tex]Slope = Speed = \dfrac{y_2 - y_1 }{x_2 - x_1} = \dfrac{x_2 -x_1}{t_2 - t_1} = \mathbf{\dfrac{Final \ position - Initial \ position }{Final \ time - Initial \ time}}[/tex]

Plugging in the values in the table, we have;

[tex]Speed = \dfrac{Final \ position - Initial \ position }{Final \ time - Initial \ time} = \mathbf{\dfrac{0.9759 - 0.8963 }{2.4 - 2.2}} = 0.398[/tex]

The speed of the car = 0.398

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