The discussion revolves around determining the instantaneous velocity at a specific time, t = 4.0 s, within the context of kinematics. Participants explore the relationship between derivatives and velocity, as well as the graphical representation of these concepts.
There is active engagement with multiple interpretations of the problem. Some participants suggest methods for calculating slopes, while others express uncertainty about their understanding. Guidance has been offered regarding the use of tangent lines and the importance of accurately reading graphs.
Participants mention constraints related to the difficulty of interpreting graph values and the need for clarity on the nature of the motion being analyzed. There is also a reference to a teacher's feedback regarding the correct approach to finding instantaneous velocity.
CWatters said:The equation for a straight line is y=mx + c where m is the slope and c is a constant.
On your graph y is distance in meters, x is time in seconds so the slope is the velocity.
The slope of the line between two points (x1, y1) and (x2, y2) can be calculated from (y2-y1)/(x2-x1) or in English.. the change in y divided by the change in x... or the change in distance divided by the change in time.
BruceW said:that would be correct if the slope was constant throughout. But in this case it is not. You need to calculate the gradient of the slope at t = 4.
BruceW said:yes, that is the right answer :)
The way I like to think of it is like this: between t=3 and t=5, the person (or whatever it is) moves 5 meters, so he moves 5 meters in 2 seconds. So during this time, his average velocity is 2.5m/s and since his velocity is the same during this time, this is also equal to his instantaneous velocity.
Fishingaxe said:This answer I got apperently was wrong.
(15 - 10) / (5 - 3) = 5/2 = 2.5. is it 2.5m/s is the wrong answer for this problem.
My teacher said to find out the Instantaneous velocity in t=4.0s I should do it like this. In the time range 2.5 s to 7.0 s, the speed is constant. Instantaneous velocity at t = 4.0 s is therefore equal to this constant speed.
We determine this by the slope. v = m / s = 1.7 m / s.
:s
PeterO said:Bruce in post #9 gave you the correct method, but (not surpisingly) could not read accurate values of displacement from your sketch graph.
While you can take the slope from the complete straight section of the graph - from time 2.5 -> 7.0, you can always take the slope from any part of that straight section.
As I said, the nature of the graph you posted makes it rather difficult to estimate actual values of displacement at various times. The real graph you were working from was presumably much easier to read accurately.