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Using a graph to find velocity and position at a said time 
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#1
Jul2111, 09:56 PM

P: 11

1. The problem statement, all variables and given/known data
Consider the plot below describing the acceleration of a particle along a straight line with an initial position of −15 m and an initial velocity of −3 m/s. A) What is the velocity at 5 s? Answer in units of m/s. B) What is the position at 5 s? Answer in units of m. 2. Relevant equations v = v_{0} + (a)(t) 3. The attempt at a solution A) Because the acceleration is obviously not constant, I added the periods of constant acceleration, from 0 to 2 seconds and 2 to 5 seconds as so: v= 3 + (7)(2) = 11 v= 11 + (1)(5) = 16  27 (answer) I'm pretty sure that I'm wrong, and I'm confused about a couple of things: 1. Is the initial velocity always going to be 3 m/s, or is it going to be the velocity that was the end point of the other equation? Because above, I got 11 m/s for the first equation  does that mean that 11 m/s will be my initial velocity for the second equation (when I'm doing the second part of the graph) as well? Or will it be 3 m/s? 2. I know this is a stupid question, but is the time variable a change in time, in other words, the final time minus the time you're starting at? Like looking at the graph, from the interval from 2 to 5 seconds when the acceleration is constant, is "t" 3 seconds, the change in time, or is it 5 seconds, the final time the acceleration is constant? I'm confused as to what to use. Once I get that squared away, I'll be able to focus on part "B" of the problem. Thanks for the help! 


#2
Jul2111, 10:12 PM

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#3
Jul2111, 10:15 PM

P: 11

v= 3 + (7)(2) = 11 v= 11 + (1)(3) = 14 Total velocity = 25 m/s 


#4
Jul2111, 10:19 PM

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Using a graph to find velocity and position at a said time



#5
Jul2111, 10:23 PM

P: 11

I'm adding the velocities because it isn't constant over the entire time, but it is constant at different intervals, so to find the velocity at a given time, I would add all the constant velocities up to that given point, right? If not, what would I do?



#6
Jul2111, 10:30 PM

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#7
Jul2311, 11:04 AM

P: 11

Thanks for the help...now for finding the position at 5s, the second part of the problem I listed above, I used this equation:
xx_{0} = v_{0}t + .5(a)(t)^2 For 0 to 2 seconds my equation was: x(15) = 3(2) + .5(7)(2)^2 = 7 Then for 2 to 5 seconds, this was my equation: x(7) = 14(3) + .5(1)(3)^2 = 39.5 However, 39.5 is coming out as the wrong answer...can anyone help me with what I'm doing wrong? 


#8
Jul2311, 10:02 PM

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For the second stage the "initial" velocity should be the velocity at the start of the stage, not the end of the stage. You've used 14 m/s as that velocity, which is the endofstage velocity.



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