From a position vs. time matching graph for a physics lab report I obtained this

Question

From a position vs. time matching graph for a physics lab report I obtained this equation in logger pro: y=0.4278 x + 0.2712. My professor wants me to use the actual variables in the equation and he is very strict with the variables and the units. Does this mean that I have to write it as: x(t)= 0.4278 t + 0.2712. Is the y intercept x = 0.2712 m ? When I write the slope m= 0.4278, what unit do I use, meter also?

I also obtained from a velocity vs. time matching graph the equation y= -0.3308 x + 1.7915. I have to do the same, therefore, do I write it as v(t) = -0.3308 t + 1.7915 ? In this case should I write the y intercept as x = 1.7915 m/s ? What unit do I give to the slope? What does that mean that the slope is negative if I was walking away from a motion detector? Please, verify the variables and the units in both equations. Thank you.

Explanation / Answer

In the first equation the unit of slope would be meter per second as it represents velocity.

In the second equation the unit of slope would be meter per second square as it represents acceleration.

The rest is OK

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