For a particular set of measurements it turns out that when the velocity is plotted on the vertical axis...?

Question

For a particular set of measurements it turns out that when the velocity is plotted on the vertical axis, and the acceleration is plotted on the horizontal axis, the result is a straight line. When a "slope triangle" is drawn for these data, it is found that the rise is 2.5 m/s, and the run is 3.2 m/s2. Which of the following is the slope?

Lab help. I have spent an hour on it and I can't figure it out.

These are the possible answers:

1.28 m2/s3
0.781 m2/s3
8.00 s
0.781/s
1.28 s
1.28/s
8.00 m2/s3
0.781 s
8.00/s

Answer 1

The slope is the change in the vertical axis with respect to the horizontal axis, so when you multiply the slope times a change in x, you get the change in y.

delta(y) =slope*delta(x)

slope = delta(y)/delta(x) = rise/run

We know that the slope must be in units of s, because that is what you would have to multiply x in (m/s^2) to get the units of y (m/s).

Since 2.5<3.5

2.5/3.5 < 1

The only answer that meets these criteria is .781s.

You can also just do the math:

[2.5(m/s)]/[3.2m/s^2] = .781ms^2/(ms) = .781s

Answer 2

Slope of a straight line is [ rise / run].

It is the same as the tangent of the angle that the straight line makes with the horizontal axis.

m = tan θ = [y/x] = [2.5/ 3.2] = 0.78125 [m/s] / [m/s^2]

= 0.78125 [m / m] * [s^2 / s]

= 0.78125 [s]
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