Believe it or not, I used to be smart. When I was younger, math and science was effortless. I also hated writing and anything connected to it.
At some point, my brain seems to have flipped around, because my sister called me yesterday to ask me for help with her son's physics homework. She gave me what I think is a trivial problem, and I couldn't figure it out.
This isn't homework help, because (a) he doesn't read Hatrack and (b) the assignment is over. But I was wondering if one of you smart folks could explain to me how this problem is supposed to be solved.
An elevator going up at .93 m/s has a downwards acceleration of .72 m/s^2 applied to it for .24 m. What's the velocity of the elevator after the acceleration?
If the acceleration was applied for .24 s, it wouldn't have been a problem. But then, my nephew wouldn't have had a problem with it, either. So is this a calculus problem? Because I've forgotten almost everything I ever learned about calc.
Posted by just_me (Member # 3302) on :
Lisa,
I don't have time to calculate it out now, but can try later.
I think the solution is to use the displacement equation: s=s'+v'*t+0.5*a*t^2. where s = .24m s'= 0 v = .93 a = -.72 and solve for t.
Then use the velocity equation v=v'+a*t v' = .93 a = -.72 t = solved for above..
hope this helps
Posted by manji (Member # 11600) on :
I don't think you need to use calculus for a constant acceleration.
Posted by Vadon (Member # 4561) on :
I'm on the bus, so I don't know how much help I'll be. I also don't have the equations in front of me to double check, but...
If I recall correctly y = y0 + v0t + 1/2at^2 if you set y0 = 0, you should have all that you need to solve the equation for T. Once you have t you can find out what the final velocity is through the equation v = v0 + at.
Shouldn't need any calculus for that. Just the quadratic formula with the first equation when solving for T.
Posted by MEC (Member # 2968) on :
if I remember correctly the formula for distance is x = 1/2 at^2 + vt + d
because the acceleration is only over .24m: x-d = .24
you then should have the formula:
-.36 t^2 + .93 t - .24 = 0
from that time you should be able to calculate the new velocity.
Edit to add: Crap, people beat me to it, I guess that's what happens when I open a lot of tabs and go back to old ones.
Posted by Lisa (Member # 8384) on :
I suck. Thanks, folks. Although... that gives two different non-imaginary times. How do I know which one to use?
Posted by King of Men (Member # 6684) on :
quote:Originally posted by manji: I don't think you need to use calculus for a constant acceleration.
You do, actually. But the integrals are very simple and so basic physics classes tend to teach the solutions, in the form of the displacement and velocity equations given by just_me, rather than going through the tedious and trivial derivation.
Posted by King of Men (Member # 6684) on :
quote:Originally posted by Lisa: I suck. Thanks, folks. Although... that gives two different non-imaginary times. How do I know which one to use?
You're getting two solutions because, if the downwards acceleration were applied continuously, the elevator would first pass the 0.24m mark, continue upwards, at some point would stop and change directions, and eventually pass the 0.24m mark again going down. So, you want the smaller time.
[Removed latter portion of post. Knock off the personal attacks, KoM. --PJ]
[ October 19, 2009, 07:00 PM: Message edited by: Papa Janitor ]
Posted by swbarnes2 (Member # 10225) on :
Can't you do this by loooking at the overall energy? Then you don't have to worry about time at all.
There's potential energy, and kinetic energy, and work, which is converted to kinetic at the end.
[Edit]
On thinking about it, I don't think potential energy matters. It'd be the same if it were train cars being pushed around.
So .5m(v0)^2 - mad = .5m(vnew)^2.
You know all of that except for vnew.
[ October 20, 2009, 07:03 PM: Message edited by: swbarnes2 ]
Posted by Lisa (Member # 8384) on :
That makes tons more sense than the other suggestions, swbarnes. And it gives the correct answer, too. Thanks.
Posted by Paul Goldner (Member # 1910) on :
My guess is he's learning kinematics, and he's supposed to use the following equation:
v'^2=v^2+2ax
Bonus: Only have to do one calculation so there's fewer opportunities for stupid calculator mistakes.
where v' is the final velocity and v is the initial velocity, a is acceleration, and x is displacement.
Posted by Lisa (Member # 8384) on :
quote:Originally posted by Paul Goldner: My guess is he's learning kinematics, and he's supposed to use the following equation:
v'^2=v^2+2ax
Bonus: Only have to do one calculation so there's fewer opportunities for stupid calculator mistakes.
where v' is the final velocity and v is the initial velocity, a is acceleration, and x is displacement.
True. That equation is the same as the one swbarnes suggested, just with everything multiplied by 2/m. You're probably right, though, that that was what they wanted.
Posted by Paul Goldner (Member # 1910) on :
Oh. Whoops. Didn't see sw's post, but yeah, his has mass and mine doesn't (which of course is mostly irrelevant since it all cancels out).
Posted by Godric (Member # 4587) on :
quote:Originally posted by Vadon: I'm on the bus, so I don't know how much help I'll be. I also don't have the equations in front of me to double check, but...
Anyone who can perform calculus on a bus, off the cuff, shouldn't be riding a bus... Unless it's by choice for environmental considerations.*
* I'm the opposite of Lisa. I was no good at all with math, but I aced every english/writing course I ever had. That said, I think those who can do math so "easily" are practically godlike. Posted by theamazeeaz (Member # 6970) on :
quote:Originally posted by Lisa:
quote:Originally posted by Paul Goldner: My guess is he's learning kinematics, and he's supposed to use the following equation:
v'^2=v^2+2ax
Bonus: Only have to do one calculation so there's fewer opportunities for stupid calculator mistakes.
where v' is the final velocity and v is the initial velocity, a is acceleration, and x is displacement.
True. That equation is the same as the one swbarnes suggested, just with everything multiplied by 2/m. You're probably right, though, that that was what they wanted.
There are four equations in kinematics you can derive by solving the second order differential equation a=d^2x/dt^2=const. Physicists who would rather start from first principles than memorize just re-derive the equations, but in HS you memorize.
They are
v=vo+at (no x) x=xo+.5*(vo+v)t (no a) x=xo+vot+.5at^2 (no v) v^2=vo^2+2a(x-xo) (no t)
and two of them were presented here in this thread.
Given that you need to know calculus and differential equations to do so, high school classes just give you all four equations without derivation, two of which you've seen here on this thread. With the equations, you don't need calculus, but one year after our 11th grade physics class, the girl who sat behind me learned calculus and was amazed at how much clearer the relation between distance, velocity and acceleration became.
Once you have the four equations, all kinematics problems become very easy. What I did in high school was write down all five variables in a column with their assigned values or whether it was the one we were looking for, circle the one that had no value, select the equation that didn't involve that variable and plug in the numbers. The hardest problems they that a physics book can throw at you involve multiple motions that have to be broken up, or make you do a calculation figure out what to plug in for a few of the five variables first, but it's always that same problem.
Your nephew will probably be moving on to 2d motion soon. Same equations, same technique, though you usually have to solve the set for one variable (x or y) to get the info you need to plug into the other set. Non existent curve-balls include the fact that there really isn't any acceleration in the X direction and you might have to do some trig to separate out the x and y velocities first, but the problem never changes.
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