# What is the displacement between $t=1$ and $t=8$? [closed]

Given the graph above, what is the displacement between $t=1$ and $t=8$?I thought this was a straight forward questions: It's supposed to be the area under the curve. Moving the time axis up to where is crosses 0, I did the following calculations: $$Area_{t=1}^{t=3.5} = 0.5 * 2.5 * 17.5 = 21.875$$. $$Area_{t=3.5}^{t=5} = 0.5 * 1.5 * 10 = 7.5$$ $$Area_{t=5}^{t=7} = 0.5 * 2 * 10 = 10$$ $$Area_{t=7}^{t=8} = 0.5 * 1 * 5 = 2.5$$

Adding the positive areas and subtracting the negative areas, I got $$\Delta{x} = 21.875 - 7.5 - 10 + 2.5 = 6.875 \space m$$ The correct answer is $8.50 \space m$. Why is 6.875 incorrect?

## closed as off-topic by stafusa, garyp, ZeroTheHero, Jon Custer, John RennieSep 16 '18 at 9:03

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• mistakes in textbooks are common, check if they used t=0 ? – PhysicsDave Sep 15 '18 at 19:52
• Welcome New contributor Art! I do believe that your question in its current form does not meet the good question guidelines found here ('check my work' questions are generally considered off-topic) so it is likely to be down-voted and closed. That said, the correct answer is, according to my calculation, 8.50 meters. Hint: the velocity for the first 5 seconds is $v = 25 - 7t$ – Alfred Centauri Sep 15 '18 at 20:44
• I concur that the answer is indeed 8.5, @AlfredCentauri, but there is a larger lesson to be learned here regarding how to eyeball points on a graph. – David Hammen Sep 15 '18 at 20:46
• @Art You are correct that this question (as do most physics problems) offers good learning opportunities. It is also true that this is a good and desirable goal to achieve. However, this does not change the goal of this site. The goal of this site is to cover and discuss physics concepts rather than become a repository of homework problems and solutions. The fact that this site has this goal and format does not mean it is against learning opportunities. This site is not a place for questions like these, even though the question itself is important for you and others. – Aaron Stevens Sep 16 '18 at 2:38
• @DavidHammen I agree with you. If the question was worded differently to ask about the best way to "eyeball" points on a graph, or how the area under a velocity graph relates to position, etc. then it would be a question that follows the guidelines for this site. However, in its current form it is just a "check my work" question, which is not ideal here. Like I say above, this does not mean the question is not useful or have good lessons in it. It just means it is not for this particular site. – Aaron Stevens Sep 16 '18 at 2:42

You did two things wrong, both being manifestations of the same problem. When one eyeballs values from a graph, one should make sure the eyeballed values are consistent with one another and with the graph, and check whether there might be better set of eyeballed values to use.

The eyeballed values in the question are

1. $v(1) = 17.5$,
2. $v(3.5)=0$,
3. $v(5)=-10$,
4. $v(7)=0$, and
5. $v(8)=5$.

The first two eyeballed values are inconsistent with a line that descends from 25 at $t=0$ to -10 somewhere near $t=5$. On the other hand, the last three eyeballed values are consistent with a line that rises from -10 at t=5 to 15 at t=10.

How good is the estimate of $t=5$ where the two lines intersect? From the descending line to the left of $t\approx5$, it's hard to tell, but from the ascending line to the right of $t\approx5$, it's fairly easy to tell that this is a very good estimate because of the nice crossings at $t=$6, 8, and 10.

If you had used $t=5$ as the point where the two lines intersect, you will arrive at the textbook's answer of 8.5.

• Are you saying that because the line left of t=5 is not perfectly crossing "good" values, it is not safe to assume v(1) = 17.5 and v(3.5) = 0, and that it why my answer is off? As @Alfred Centauri pointed out, maybe using the actual velocity function and integrating would be a safer bet? Is it even possible to use areas to determine the displacement in this graph? – Art Sep 16 '18 at 0:47
• @Art - If you assume $v(3.5)=0$ and assume that the two curves are lines, then you should have used $v(1)=17\,6/7$ and $v(4.9)=-10$. This choice would have affected the right hand side calculation as well because that places the intersection of the two lines at 4.9 rather than 5. – David Hammen Sep 16 '18 at 11:34

You need to notice that the acceleration from t=0 to t=5 is constant, and given by the slope of the line. And the acceleration from t=5 to t=10 is constant, also given by slope of the line. Given that, you have the ability to get the specific equation for the location of the particle as a function of time. That's just $$x(t) = x_0 + v_0 t + \frac{1}{2} a_0 t^2$$.

So you can easily solve this equation for the case from t=0 to t=5 with one acceleration. Then you can do a very similar thing from t=5 to t=10 with a different acceleration.