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Equation of motion question 
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#1
Sep1711, 09:02 PM

P: 651

s = ut + (1/2)at^{2}
if s = 0, a = 2u / t so since acceleration is change in velocity over time, so change in velocity = 2u why is the change in velocity given as 2u??? 


#2
Sep1711, 09:35 PM

PF Gold
P: 3,080

The key is to understand what equations like that mean. A graph can be very useful, so see if you can make a graph with t on the x axis and s on the y axis, and use that formula to get s(t). Can you see that if you set s=0, you are not making a general statement about s, you are asking for the times when s will be zero? That means you want the times when the object, whatever it is, has returned to its starting point, and there are only two times that will work. One of them is trivial the t=0 when the problem begins. The other time is 2u/a, which comes at an earlier point when the object was actually moving backward. But when you solve that expression for a instead of t, what you have is a formula that asks "what acceleration do I need such that the particle will return to the place where it had velocity u at a time t later." That's kind of an awkward thing to ask for, but the equations allow it you just have to realize it is not a general statement about all a and all u and t, it's asking for a special "a" given some pair of "u" and "t" of interest.



#3
Sep1711, 09:38 PM

P: 329

Also note that u is the initial velocity. The only way the ∆v can be the negative of 2v_{i} is if ∆v=2v_{i}=0 (which also makes a=0). 


#4
Sep1711, 09:48 PM

P: 651

Equation of motion question
oh so , if i want s = 0, then my possible times are 0 and 2u/a ?
because t= 2u/a nicely satisfy the equation of motion, letting s =0. but we don't consider this negative time because we haven't started timing our experiment? i see... thanks! 


#5
Sep1711, 09:53 PM

P: 329




#6
Sep1711, 09:56 PM

P: 651

if v_{f} = v_{i}, then ∆v = v_{i}  v_{i} = 2v_{i} ?? 


#7
Sep1711, 10:27 PM

P: 329




#8
Sep1811, 08:09 PM

P: 651

from s = ut + 1/2 at^{2} 0 = 5(t) + 1/2 (9.81)(t^{2}) t = 0 or 1 so this time is for the total journey right? so the time to reach the peak height would be 1/2 s? then for it to come down , another 1/2 s? 


#9
Sep1811, 08:14 PM

P: 329

Problems like this are interesting when the initial position is higher than the final position (like tossing a stone vertically upward and then letting it fall into a well). You still get two values for time (when solving for the landing in the well). I was always told to throw out negative time solutions... do you see what the other root means in a problem like this? 


#10
Sep1911, 08:20 AM

P: 651

hmm, i am not sure, i was also taught to ignore ve times.
if i have this setup, taking upwards as +ve s = ut + 1/2 at^2 5 = 10t + 1/2 (10)t^2 so t = 2.41 or 0.41 but these are equations. equations don't care how you obtain s = 5 , as long as you get it, it doesn't care which route you take? so i would guess 0.41s is the time it takes for the particle to travel downwards? since my ut value has effectively become ve, it is as good as saying the particle has downwards direction with positive time? by the way, when i think about the stone throwing upwards and then falling downwards, i find it weird that it actually takes the same time for going up and coming down. although the equations points to that fact, if i were to shoot a marble upwards at 1000m/s, it would keep going up and cover a huge distance, but on the coming down part, it is only free fall, at 10m/s^2. this doesn't seem intuitive at all? 


#11
Sep1911, 03:17 PM

PF Gold
P: 3,080




#12
Sep1911, 08:06 PM

P: 329

A stone is launched from the ground level with an initial velocity of 10 m/s upward. It rises to its peak and then comes down in a 5 m deep well. How long does it take to land in the well? The 0.41 s is the time that the stone would have been launched from 5 m to reach the ground level of the problem moving 10 m/s upwards. (It might help to visualize this as a parabola. When you look at the problem starting at ground level, one leg of the parabola is cut off. If you go back to 0.41 s, you get the entire parabola. For me, thinking about the math describing a body being dropped from 10 m helps. Given the initial velocity of 0 m/s, the acceleration of g and the distance, the time is determined. The final velocity is 20 m/s. Now consider throwing the body to a height of 10 m. The acceleration and distances are the same. There are the velocities of 0 m/s and 20 m/s at the ends of the trip. This makes me feel the time being the same makes sense. 


#13
Sep2311, 08:31 PM

P: 651

thanks every1



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