Determine from the time / height graph the speed at point

[moenste]

Homework Statement

Determine from the left graph the speed at t₁.

Answer:
Draw the tangent (right graph) at t₁ and determine the slope. v = 1.4/0.30 = 4.7 m/s

2. The attempt at a solution
The slope is (Y2-Y1)/(X2-X1). t₁ coordinates are (0.24-0.25, 0). In order to get the answer I should draw a tangent line exactly in the (0.54, 1.4) coordinates. How is it possible? As I uderstand the tangent line should be as close to the curve as possible, before crossing it. I printed out the left graph and I get like (0.58, 1.4). So the answer is 4-4.3 depending on the 0.24-.25. So my questions: Am I plotting the tangent line right? (the closest as possible to the curve) If not, how should I plot it? Where is my mistake, so I can get the 4.7 m/s answer?

Thank you all in advance.

 

[phinds]

I think you're doing everything right, it's just that it's impossible to get a very accurate answer under these conditions. I got 3.5 at a quick eye-balling but after looking again, I can support your answer. 4.7m/s seems at bit of a stretch and I can only conclude that the guy who came up w/ it drew HIS tangent a bit off.

 

[moenste]

I think you're doing everything right, it's just that it's impossible to get a very accurate answer under these conditions. I got 3.5 at a quick eye-balling but after looking again, I can support your answer. 4.7m/s seems at bit of a stretch and I can only conclude that the guy who came up w/ it drew HIS tangent a bit off.

Thank you. And the logic that tangent lines should be drawn as close to the curve as possible, but not cross it, is correct, right? And in general: draw the tangent line as close as possible to the curve, use coordinates of the point (in my case t₁) and the coordinates where the line goes off the graph (at 1.4 meters) and using the coordinates get the slope.

I wouldn't be so worried, it's that the problem is from the past exams and this is the answer, so I guess it's right. But still I guess if the tangent line is drawn correctly and the coordinates are taken right, I don't think it's a big problem if the answer is slightly different. After all, the graph is really bad, can't tell what's the t₁ coordinates (0.24 or 0.25 or else) and same for the second point, the x coordinate is also hard to measure.

 

[BvU]

Yeah, I can easily get 1.4/0.28 = 5 and still claim that's a good tangent :

I agree with you that grading this should allow a considerable margin. Also I think the exercise is a bit unfair asking for the tangent at an end point.

 

[phinds]

Thank you. And the logic that tangent lines should be drawn as close to the curve as possible, but not cross it, is correct, right?

Right. Ideally, the tangent should touch the line at exactly one point but not by crossing. The curve should move away from the tangent line in each direction from the point of contact. I agree w/ BvU that asking for the tangent of an end point on a graph is unreasonable.

 

[BvU]

Overdoing it, I admit.
Here's a picture for a parabola ##v_0 t + {1\over 2} at^2 ## with ##v_0 = 4.9## and ##a = -9.8 \quad ## (##t_0 = 0.24##)
Not bad, but at t = 1.08 a bit low.

 

[Ray Vickson]

Overdoing it, I admit.
Here's a picture for a parabola ##v_0 t + {1\over 2} at^2 ## with ##v_0 = 4.9## and ##a = -9.8 \quad ## (##t_0 = 0.24##)
Not bad, but at t = 1.08 a bit low.

View attachment 88712

Nice graph. What package did you use to make it?

 

[BvU]

The epitome of unstructured programming. It's called Excel.