Question

I'm doing a write up on a lab for the speed of a dune buggy. I'm making a line graph (distance vs time) with time as the Y axis because its the dependent variable. Anyway, what would the equation be in terms of speed? (I used meters and seconds)

Answer 1

Speed is calculated using
\[v=\frac{ \Delta d }{ \Delta t }\]I believe your choice of dependent variable is a mistake. Choose time as the independent variable and distance as the dependent variable. You may then recognize the above equation as the rise of your graph over the run of your graph, in other words, the slope of your graph. Speed is the slope of the d-t graph.

Answer 2

Why would time be independent if the time depends on how long it takes to get to a certain distance? also, my teacher gave us this formula if it means anything. I mightve recorded it wrong
d= (m/s)t+(headstart error)

Answer 3

The distance the dune buggy travels depends on the amount of time you let it run. If you allow it to run for a short period of time, it will not travel very far. If you allow it to run for a longer time, it will travel a greater distance.

The equation given by your teacher is simply a rearrangement of the one I provided, i.e.
\[\Delta d = v \Delta t\] I don't know what is meant by headstart error. It must be something specific to your experiment.

Answer 4

There wasn't a set amount of time, but a set distance
and is that displacement= speed(time) (just to clarify, this is my first week of physics)

Answer 5

Correct, it is distance = speed x time.

Answer 6

You'll have to explain your experiment a little more. You mentioned the dune buggy travelled a set distance. What did you vary for each trial?

Answer 7

Okay so, we did 12 trials. 1st, we timed how long it took for the dune buggy to travel .25m. then, .5m and so on adding 0.25m each trial until 3m

Answer 8

OK. If I were doing it, I would still choose time as the independent variable and distance as the dependent variable. You are still allowing the dune buggy to run for longer periods of time in order to cover greater distances. I'm assuming your dune buggy moved with a constant speed (battery powered?). If your goal is to calculate the speed of the dune buggy, it is simply the slope of the graph (if you plot time on the x-axis ans distance on the y-axis). That's how we do it in my class.

Answer 9

Ah okay, thank you. I'm still confused about the formula though...how do you know what goes into each space? I'm so lost

Answer 10

Assuming you get a straight line, or very nearly a straight line, slope is calculated by choosing two points on the line of best fit \[\left( x_1,y_1 \right) \text{ and } \left( x_2,y_2 \right)\] and using the formula for slope\[m=\frac{ y_2-y_1 }{ x_2-x_1 }\]

Answer 11

I can calculate the slope on my calculator, but I was wondering about the speed x time. What speed and time do I use?

Answer 12

and is distance just the same as y? I dont put a number there?

Answer 13

I'm sorry, I don't understand. What are your objective in this lab? I've assumed the goal was to determine the speed of the dune buggy. Is it something different than that?

Answer 14

no, that't it. I just have to do a full write up on it and I'm stuck at the analysis, which includes a graph and an equation for the line on the graph. For some reason , the graph did not come out as perfectly linear. I can send a picture of the graph if its helpful.

Answer 15

Experimental results are rarely perfect, but I assume your graph came out to be ROUGHLY linear. You need to draw the line of best fit, the straight line that best suits the data. It may or may not pass through any of the data points. Then, to calculate the speed of the dune buggy, choose two points ON THE LINE OF BEST FIT and calculate the slope. This slope will be the speed of the dune buggy,

Answer 16

You might use the formula\[v=\frac{ \Delta d }{ \Delta t }=\frac{ d_2-d_1 }{ t_2-t_1 }\]

Answer 17

Ohhh okay. I'm going to put the exact graph and the line of best fit graphs on the write up because this is my first lab and I'm not sure what my teacher wants.
Is that formula for slope?

Answer 18

You only need one graph. The line of best fit should be drawn right on the graph with your data points. That way, it's easy to see how well they fit together. And your write up should include some discussion about why your data isn't exactly linear, experimental error and all that stuff. I feel your pain, lab reports are challenging, and different teachers have different expectations.

Answer 19

Oh, and that formula is for slope

Answer 20

Thank you so much! So the equation is for the line of best fit and NOT the raw data?

Answer 21

Correct. Choose two points on the line of best fit, not the raw data \[\left( t_1,d_1 \right) \text{ and } \left( t_2,d_2 \right)\]and apply the formula.

Answer 22

Could I use linereg on my calculator? In that case, I got y=2.3969x+0.285 as the equation but I'm still not sure how to convert that to physics terms

Answer 23

So in which case the slope would be 2.3969, correct?

Answer 24

That's correct. Therefore, that's the speed of the dune buggy. Theoretically, the line should pass through the origin (at t=0 seconds, the distance should be 0 meters). The fact that it doesn't (it has a y-intercept of 0.285) is not unusual. It's an indicator of experimental error and you should discuss this in your write-up. Consider things like human error in taking measurements, quality and calibration of equipment, variation in the speed of the dune buggy due its construction, etc.

Answer 25

Oh so the .285 is the headstart error he was talking about! So y=2.3969x+0.285 would be the equation but I would change it to t=2.39d+.0.285? (well, I still need to switch the variables so that t is the independent, but thats the gist?)

Answer 26

You got it!

Answer 27

Yay!!! Thank you so much for sticking with me on this, I know I'm not the brightest of kids, especially so late at night haha. I really appreciate all of your help!

Answer 28

You're welcome. I think your first lab write-up is going to be great! Good night.

Answer 29

Good night