So, I just finished my midterm for Physics, and I understood every problem except this one. The test is over, so I don't need the answer, really, but I would like to know what it should have been. Here's the question.
The movement of a particle in an xy plane is given by the equation:
r = (5(t^2)m/(s^2))i + (6m-(4(t^3))m/(s^3))j
a. Give the Y velocity at t=3.
b. Give the X acceleration at t=3.
c. Find the magnitude in the x-direction of the displacement vector of the particle between t=3 and t=6.
The whole question completely stumped me. Anyone know how to solve it?
Wait... did you stick units into that? O.o
Without units, is it this?
r = (5(t^2))i + (6-(4(t^3)))j
Nope, the units were already there. That's why it was so confusing. Without the units, it would have been a very simple problem.
lol just ignore the units they aren't relevant to the actual math as long as you give the answer in correct units
Quote from: Ryexander on September 24, 2010, 02:52:06 am
lol just ignore the units they aren't relevant to the actual math as long as you give the answer in correct units
Those are all variables.
Quote from: Ryexander on September 24, 2010, 02:52:06 am
lol just ignore the units they aren't relevant to the actual math as long as you give the answer in correct units
I don't think that's right. On the contrary, my professor told us earlier this year to, as Dio said, treat the units as variables. That's probably the point he was trying to make with this problem.
that doesn't even make scene... units have no value... well seems I can't help then, perhaps next semester if i have physics.
I'm guessing we were supposed to somehow use dimensional analysis to figure out what state in which each vector component was given.
:( It makes me sad that instead of teaching physics, your professor is dicking around with units in an equation. Nobody puts the units into an equation at different points, they just put the final units in at the end: r = ((5(t^2))i + (6-(4(t^3)))j) m, and that's the end of it.
Agreed. Yes, unit are important, but it's more important that you know what's going on. If you don't know, then knowing the units won't help you while vice versa it's not such a bad situation. If you don't know which unit it is, just look it up. You already have the correct result. :/
I'm not sure I get it, the x is acceleration and y is velocity? Might be easier to imagine two graphs, one for acceleration vs time, one for velocity vs time.
I think you would have to plugin 3 for t and solve for m/s and m/s^2, somehow. But m/s^3 is like double acceleration, I've never seen that before. Or I'm confused with something else...
That's always the problem with questions that rely on being able to answer the preceding. Being able to get the magnitude you need to do the complex math beforehand. Lol, I would've skipped that question as soon as I've seen it...
Quote from: poxysmash on September 24, 2010, 04:08:03 pm
I'm not sure I get it, the x is acceleration and y is velocity? Might be easier to imagine two graphs, one for acceleration vs time, one for velocity vs time.
I think you would have to plugin 3 for t and solve for m/s and m/s^2, somehow. But m/s^3 is like double acceleration, I've never seen that before. Or I'm confused with something else...
That's always the problem with questions that rely on being able to answer the preceding. Being able to get the magnitude you need to do the complex math beforehand. Lol, I would've skipped that question as soon as I've seen it...
m/(s^3) are the units for jerk. Apart from that, your comment pretty much followed my exact line of thinking.
t is in seconds, if that helps. so t^3 m/s^3 just comes out to meters
r = ((5(t^2))i + (6-(4(t^3)))j) m
v = 10t i + 12t^2 j m/s
a = 10 i + 24t j m/s^2
a. vy(3) = 12(3)^2 = 12(9) = 108 m/s
b. ax(3) = 10 m/s^2
c. rx(6) - rx(3) = 5(6)^2 - 5(3)^2 i = 5*36 - 5*9 i = 180 - 45 i = 135 i
Quote from: winkio on September 24, 2010, 07:39:47 pm
t is in seconds, if that helps. so t^3 m/s^3 just comes out to meters
That's what I thought, too, but wouldn't that mean that the position was constant?
Quote from: Aqua on September 24, 2010, 07:52:53 pm
r = ((5(t^2))i + (6-(4(t^3)))j) m
v = 10t i + 12t^2 j m/s
a = 10 i + 24t j m/s^2
a. vy(3) = 12(3)^2 = 12(9) = 108 m/s
b. ax(3) = 10 m/s^2
c. rx(6) - rx(3) = 5(6)^2 - 5(3)^2 i = 5*36 - 5*9 i = 180 - 45 i = 135 i
That way simply changes the units to meters, when they aren't, according to the initial problem. If you're trying to follow winkio's logic, the t's on top and the s's on bottom would cancel out. You left the t's in the problem, and just removed the s's. I'm pretty sure that's incorrect.
Then put the s's back in.
They'll cancel out when you plug in 3 seconds and still get what I got.
Quote from: Aqua on September 24, 2010, 08:26:52 pm
Then put the s's back in.
They'll cancel out when you plug in 3 seconds and still get what I got.
<3
On a related note, I just got this email from my professor:
Quote
As we have graded Exam 1 is clear that a large number of you did poorly on Problem 2 on the second part because you didn't understand the problem. Part of the problem was to understand it, but I would still like to give some of you a second try at it. So, if you are one of the CID numbers in the following list you may try this problem again. All of you on this list received a score of 8 out of 15 on Problem 2. You may either just accept this score, or you can try it again for a maximum of 12 points. The testing center doesn't want 100 students descending on them when they are not planning for it, so I will have to have you take it in some other way.
The first time you can take it is in our classroom from 10:15 to 10:45 or from 2:15 to 2:45 tomorrow, Thursday, September 30. I don't care which section you are in, if you can come sit in the classroom during this time and retake Problem 2, please come do so. If you can't do it at one of these two times you will have to come see me to arrange to take it sometime on Friday. The tests will be handed back on Monday, so if you want to retake this problem you have to do it either Thursday or Friday.
Here is the list of CIDs who can retry this problem:
018 026 042 046 056 085 092 095
107 116 125 140 147 157 159 161 162 176 191 199
209 210 224 246 249 257 260 261 275 280 290 291 298
306 307 316 335 344 346 356 360 376 377 378 393 399
404 406 411 416 425 428 447 456 463 472 479 481 484 494
501 410 519 522 530 539 542 543 554 556 558 562 574 575 583 591
602 610 622 633 645 655 677 679 686 689 692 695
721 737 745 746 750 756 770 779 781 784 795 797
813 820 821 827 829 831 845 856 860 891 898
918 920 923 934 943 955 975 976 977 978 986
One of those is my number, so now that I more fully understand the problem I should have an easier time with it.
So basically do what I wrote? :P
Quote from: Aqua on September 30, 2010, 03:02:40 pm
So basically do what I wrote? :P
Pretty much, though of course he changed the numbers this time around, but the concept was the same. I think I did well this time around.
Awesome :)