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Velocitytime graph question 
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#1
Dec1706, 04:13 PM

P: 6

1. The problem statement, all variables and given/known data
How far does the runner whose velocitytime graph is shown in Fig. 230 travel in 10 s? 2. Relevant equations x = v0t + 1/2at^2 3. The attempt at a solution x = 10s (0) + 1/2(8m/s)(10s)^2 x = 400m 


#2
Dec1706, 04:18 PM

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P: 3,033

The x = . . . equation you wrote is relevant to the problem, but there are other relationships involving velocity and time that are more directly useful in this problem. The fact that you are given a graph suggests that the problem is looking for one such relationship in particular. Can you say what it is?



#3
Dec1706, 04:20 PM

HW Helper
P: 3,224

Hint: the displacement from t = 0 to some time t1 equals the area under the v(t) graph from the points t = 0 to t1.
Edit: late. 


#4
Dec1706, 04:21 PM

Mentor
P: 8,297

Velocitytime graph question
There's an easier method to use here, since you are given a graph. Do you know a way to find the distance from a velocity time graph?
edit: haha, even later! 


#5
Dec1706, 04:44 PM

Sci Advisor
P: 987

The whole point of the question is to let you look at the graph, make your own formulas, and come up with an answer. Just looking at the units of velocity and time, you can say that multiplying them together will give a unit of distance. The first basic shape you can see in that graph is a triangle where the runner goes from 08m/s in 2 seconds. What's the formula for a triangle? (1/2)(height)(base), which would be (1/2)(Vf  Vi)(t). If you multiply that by t/t (which is 1), you get (1/2)(a)(t^2). d = (1/2)(a)(t^2) is one of the basic equations you are given in physics, and now you know where it comes from. 


#6
Dec1706, 05:36 PM

P: 6




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