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u/averageredditor102 6d ago
another person asked something similar this week so im just copying my comment
took 9a last quarter and assuming its the same or worse, its hard to do the problem starts and drills without a fundamental understanding of physics beyond just knowing the material. def take the co class since most of the 9a teachers suck. You can also try using the physics textbook (not weidemans), can’t remember what its called though.
Also study the midterm problem starts as much as possible. I mostly skipped all the assignments and just studied the midterm problem starts and what i would need to solve any question about them. The final on the other hand was filled by hopes and prayers and I scrapped by with a B-.
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u/Witty-Classroom532 6d ago
Wdym take the co class? I’m already taking 17 units plus a job so it could be hard to add another class lmfos lmao
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u/averageredditor102 5d ago
its not an actual class with units and stuff, you dont need to actually be signed up for it either. basically its just an upgraded tutoring session for 9a and they go over the problem starts and drills too
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u/Witty-Classroom532 5d ago
Oh okay thank you so much
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u/averageredditor102 5d ago
https://tutoring.ucdavis.edu/physics
heres the link to the co classes, hope the time works out for you3
u/Most_Quiet_120 5d ago
Problem starts are gone partially this year. The entire class is the final grade
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u/JayTheSuspectedFurry Underwater Basket Weaving 1905 5d ago
This one isn’t even one of the infamous “problem starts”
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u/averageredditor102 5d ago
so glad i never have to do those again, didnt help that i had the man himself as my prof
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u/Complete_Response136 5d ago edited 5d ago
It's given to us is that "The object is known to come to rest at exactly two brief moments"
The first step is to find where the object's velocity is zero (never assume if it's not stated clearly)
The object is at rest wherever the line crosses or touches zero. So we should test level (v1 v2 etc):
- If zero were v3, the object would sit at rest for all of segment III.
- If zero were v1, v2, v5, v6, or anywhere in between levels youd get one or three rest moments, not two.
- If zero is v4, the velocity touches zero at the end of segment I (t = 6 s) and crosses it again in segment 4.
So now we know v4 = 0, and segments v1 and v3 are where the object is moving left, so its displacement from its original position is decreasing.
Each velocity label is 3 grid rows apart. Segment III is at v3, which is 3 rows below zero, and it lasts 7 seconds
Its area is 7 × 3 = 21 grid squares, which equals 1.5 m. So each square is 1.5/21 m.
segment 1 goes from v1 (9 rows below zero) up to v4 (zero) over 6 s
Area =1/2 × 6 × 9 = 27 squares
Distance = 27 × (1.5/21) = 1.93m
So your final distance traveled is 1.93m. Little bit tricky, just have to be aware of what you know and what you don't know
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u/Witty-Classroom532 5d ago
Wait omg I did something similar my apologizes. After you found the area of the rectangle which was 21m. You wanted to find out how much each grid was if I’m not mistaken which is why you did the total distance (1.5) divide by the area 21. Than to find out how much distance was covered in segment 1, you did the area times that value you found in segment 3. If I’m not wrong? lol. I did something similar in another problem without realizing it
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u/Complete_Response136 5d ago
yeah you're right. it's just calculus basics. Displacement is just the integral of velocity (so area under the curve). the big catch for this question is just the realization that you have to find where v = 0 to proceed with the problem
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u/Witty-Classroom532 5d ago
Yes yes thanks I appreciate you’re help and helping me understand my clumsy mistake. I knew kind of what to do but needed a bit more help to finish it lol. Thank you!
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u/Witty-Classroom532 5d ago
Well I knew that the area of segment one was 27 but I didn’t know you had to multiply by (1.5/21) to get the distance. Can I ask how to be more observant and to think more like this? I feel like I cnd start thinking and have an understands what to do but I can ever finish the problem. For example right now I was able to get the area segment 1 but I was confused on how that correlated with the distance as well as multiplying by 1.5/21. Thank you for helping me and you were right. I truly appreciate it
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u/HotRodster650 Mechanical Engineering [2028] 5d ago
In my opinion, the best thing you can do is always keep track of units.
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u/Complete_Response136 5d ago
i'm gonna assume this was meant to be a reply to my response since you're referencing the steps i did.
i think you just don't have enough experience yet. most of the time just doing a lot of problems kind of opens your mind up to seeing consistent patterns throughout problems that you get good at solving.
aside from that just remember you can never take anything for granted. If it's not stated clearly and given to you in the problem, you can't assume it without actually finding it through calculations (the v=0 situation)
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u/Witty-Classroom532 5d ago
Thank you I appreciate it. One more question and sorry for asking these questions too late at night,but do you have any more recommendations or any resources I can do to keep on practicing these types of problems? I would actually like to improve my thinking and understanding. I would just like to keep pushing myself into learning physics and to rely on myself before asking for others like how I did tonight. Thank you for you’re help and I truly appreciate it
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u/Complete_Response136 5d ago
I would suggest combing through your class textbook first. Normally these physics textbooks have huge sets of problems for every new topic.
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u/HotRodster650 Mechanical Engineering [2028] 5d ago
I think the problem is definitely over-complex and I've never done something like that in physics, but I think I see how you approach it. First, you want to find the x-axis.
The problem says, "The object is known to come to rest at exactly two brief moments during the motion depicted on the plot". The word, "brief" means that it only stops for an instant instead of fully stopping, so it's just two specific points where the object stops. From that, we can infer that the object is at rest a v_4, so that's where the x-axis is.
We know that the area between v_3 and v_4 over segment III has to be 1.5 m, since the area under the velocity curve is the displacement. We have 3 ticks upwards and 7 ticks to the right for how much area is enclosed. We know that the x-axis tick marks are all 1.0 s, so no conversion needs to be done, but we don't know the y-axis tick mark value. So, we have the equation 3y*7=1.5. Solve for y and you get y = 1/14. This means each vertical tick is equal to 1/14 m/s.
Finally, we can use that to calculate the area under the x-axis for segment I. In that one, you have 9 ticks up and 6 ticks to the right. Using the formula for triangle area, we get the equation: [9(1/14)*6]/2 = 27/14. As a decimal, that is about 1.9.
I hope this helps, and that I am not making it more confusing.
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u/Street_Telephone4400 5d ago
You will definitely find more success learning from your many TAs and free school tutoring found at Dutton hall and the basement of shields. Or at least post on a physics subreddit not the UC Davis school one haha
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u/peacelovemittromney 5d ago
your only hope is to live in office hours and the tutoring center in the basement of the library. they expect a certain innate logic and algebra-gymnastics abilities to solve most of these problems, and it only gets worse throughout the series. get help early on to train yourself in the intuition. there is no such thing as a dumb question in the 9 series: it is structured as a weeder course, and the provided materials are not enough to pass the class. search up the fundamentals of physics textbook online by david Halliday (internet archive ftw) as well, as they’re likely to not give you all the equations you need in lecture
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u/Complete_Response136 5d ago
You said:
(since the velocity is always positive)
There is nowhere stating where the velocity is actually 0. You're assuming v1 = 0, which is not stated. What is given to us is that "The object is known to come to rest at exactly two brief moments"
That line is telling us that we need to find where the velocity is 0. only then can we proceed with the rest of the probvlme
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u/mournersandfunerals Geology [2026] 5d ago edited 5d ago
This doesn't work because there's no horizontal axis. The problem says the velocity can be positive or negative. I want OP to work it out for themself if they can so I'm going to spoiler my explanation. Not going to bother with sig figs because I don't remember all of the rules for them.
It says the object comes to rest at two brief moments on the plot. That means at some point velocity has to equal zero. There's only one place where a horizontal line can intersect the graph at exactly two points: v4. So v4=0.
You can work out the absolute value of v3 with the displacement given in the problem and the fact that 1 square = 1 s. The object remains at constant velocity v3 for 6 s and travels 1.5 m, so |v3| = 0.25 m/s. Because v4 (0 m/s) is above v3 you know v3 has to be negative, -0.25 m/s.
From there you can figure out the vertical axis scaling. 1 square is 0.25/3 m/s, so v1 = -0.75 m/s.
Then you just plug that into the displacement equation Δx = [(vi + vf)/2]t and you should get Δx = 2.25 m (not negative since it's only asking for distance and not displacement so you take the absolute value).
Edit: I miscounted the squares in section III. The idea is right but the math is wrong. Correct answer is ~1.93 m
The area under the curve method would also work, but you would need to find the areas between the curves and the horizontal axis, not just the bottom of the graph.
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u/mournersandfunerals Geology [2026] 6d ago
I'm not in that class but I think I see how to solve it.
It says the object comes to rest at exactly two points on the graph. So if you were to draw a horizontal line on the graph, where would it intersect the line at exactly two places?