How Do You Analyze Motion Using V-T Graphs?

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The discussion focuses on analyzing motion using velocity-time (V-T) graphs, particularly through a series of calculations involving displacement, velocity, and acceleration. Participants emphasize the importance of correctly calculating displacement during each time interval and sketching the corresponding V-T graph. It's noted that the initial velocity of the cart affects the total displacement, as the cart was not at rest at time zero. The use of the equation s=ut+(at^2)/2 is suggested to find the initial velocity, which can then be incorporated into the V-T graph for accurate area calculations. The conversation highlights the challenges of understanding motion analysis and the significance of initial conditions in kinematic equations.
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Homework Statement



distance (cm) 0.0 5.0 15.0 30.0 50.0 75.0 105.0 140.0
time (s) 0.0 0.1 0.2 0.3 0.4 0.50 0.6 0.7

a. calculating the displacement during each time interval. Add to the above table of data.

b. calculating the velocity of the cart during each time interval. Add to the above table of data.

c. sketching a velocity-time graph.

d. calculating the acceleration of the object from the velocity-time graph.

e. Using the velocity-time graph to calculate the total displacement of the cart.


Homework Equations





The Attempt at a Solution



http://www.freewebs.com/dazed42/phys.htm
 
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Ok, you've used the right method, and therefore the answer will be correct as long as you've not made any arithmetical errors. The reason for the displacement not equalling the total distance measured will be the same as the previous question, namely that the velocity of the truck at t=0 was no zero.

So, you can use the equation given in the last question to calculate the initial velocity. Recall s=ut+(at^2)/2. Put s=140, t is the total time (0.7s) and a is the acceleration you have calculated.

Really, for the question, you just need to state the the truck was not at rest at t=0, but if you want to check, you can calculate u from the above equation, then add it to each value of v on the graph, giving you a new graph, for which the area underneath will add up to 140cm
 
Thanks buddy. From the phrasing of the question I was given the impression that the initial velocity could not be zero. (the cart is at rest before 0 and released at 0, so there must be some instantaneous acceleration)

Much thanks for all the help, I'll be glad to put this behind me now. (Until finals.. whee)

peace,
Dazed
 
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