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???
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.
I think you are being a little fast and loose with your reasoning. First note that if S=0 it could be because u and a are zero. The other alternative is that t=0. Your algebra rules out that latter possibility(since you are dividing by t). 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).
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!
Also keep in mind that s is really the change in position. That is handy when working with a situation where a body returns to its original position (like when vertically tossing a ball). In that case ∆s=0.
Interesting. That would work for something like a ball launched vertically at 5 m/s. It returns to the launch point at -5 m/s, regardless of the values of a and t.
hmm, in this example, i can find my a and t values right? assuming a = 9.81 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?
Correct. 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?
hmm, i am not sure, i was also taught to ignore -ve times. if i have this set-up, 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?
If you neglect air resistance, the motions are symmetric because the -10m/s^2 is there the whole time, both up and down. (Air resistance is really important in practice with the numbers you are using though!) It sounds like you're problem is that 10m/s^2 doesn't sound like much, if you shoot the marble at huge speeds, but remember the faster you shoot it, the higher it goes, because it will take that -10m/s^2 a longer time to stop the upward motion. Since it took longer to stop it, it will have longer to speed back up on the way down, so gets just the right amount of time to return to you at the speed you shot it up at.
It looks like that is the answer to this question: 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. I agree: it is counterintuitive. I think that is because we do not have very good intuition about uniformly accelerated motion. 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.