Thread: How much time did it really take?

Since I joined this forum I've been asking stupid questions. This is another one :

We are all familiar with this: s=vt+(at^2)/2

Now if the initial velocity, the acceleration, and the displacement are known, the time taken can be calculated through this:

t = [-v(+-)sqrt(v^2+2as)]/a ( note: +- means plus or minus )

Now we have two values of t!

I wanted to know what this would look like in the v-t diagram but never got around to it.
Please help explain this!

Hmm... looks like my question is kind of hard, so far none got an idea?

Here's the thing. while the equation



Gives two possible answers, only one has any physical meaning.

One will will give you a negative value for t, which is nonsensical.

You get the same kind of thing if you were to solve



for r,

you get



and of course, only a positive value for r makes physical sense, so that's the one that's correct physically.

Yes I know that, but it has to have some meaning! How can it have no meaning!

My guess is: If an object was moving from the origin to the destination point, it takes t1 seconds to get there; what the negative t2 insinuates is a reverse motion, or something like that.

The problem is that when you use your equation to get your result, you are assuming a model of the situation. Our models have certain limitations. Your model of the motion of the object doesn't take into account the environment. It assumes that the motion of the object is described precisely by this equation. But of course it wasn't because presumably somebody dropped the ball, or it fell off some surface, at which time the description of the motion was different.

This model here is that of a parabola. A parabola crosses the x axis at two places, if you consider the x intercepts to be the object at ground level then the two values of t correspond to the object leaving the ground and returning to the ground. Its just often this model is applied where t = 0 falls within those two intercepts. Sorry if this was confusing.

Originally Posted by TheObserver

The problem is that when you use your equation to get your result, you are assuming a model of the situation. Our models have certain limitations. Your model of the motion of the object doesn't take into account the environment. It assumes that the motion of the object is described precisely by this equation. But of course it wasn't because presumably somebody dropped the ball, or it fell off some surface, at which time the description of the motion was different.

This model here is that of a parabola. A parabola crosses the x axis at two places, if you consider the x intercepts to be the object at ground level then the two values of t correspond to the object leaving the ground and returning to the ground. Its just often this model is applied where t = 0 falls within those two intercepts. Sorry if this was confusing.

Um... no it's not a parabola. The v-t diagram is a straight line. Where do you get a parabola? Keep in mind that this is for rectilinear motion.

Yeah sorry s-t graph is a parabola, the v-t graph is just a line. Not sure what the v-t line has to do with the original question so I assumed you meant the graph of position.

The v-t equation is just v = v0 + at. (v0 being the initial velocity)

Hmm... you're right, I should be analysing the s-t diagram, I should then be able to work out what the negative value's meaning is.

Good question, thanks!

Ok the +- t comes from the two intersections on the x axis. But still, what does it mean? I know the algebraic meaning, the geometric meaning, but... using everyday language... what is it! Sorry for being persistent.

Well this equation describes the motion of an object being acted on by a single constant force. For example this could be the vertical motion of an object being acted on by gravity alone, using the approximation that gravity is constant at all heights. If we drop the object from some height s0 then solving this equation for t will give you the time that the object reaches the new height s supposing that this height s is a lower point than s0. Obviously if we dropped the object from a point lower than s, it will never reach that point. In this case you will find a negative number under the root sign.

If s is lower than s0 then you will find two solutions. The positive solution will be the correct time. What would the negative t value here be?

Well imagine instead that you threw the object up in the air from the ground until it reached a height s0. As soon as it reached s0 you start your timer. This is now t = 0 and you are using the identical equation from before to model the situation. It takes the exact amount of time to reach the point s as if you had dropped it from s0 as before. But even better the model holds right from the time you threw the object up so that the negative solution here would correspond to the time it reached the height s on the trip upwards. Here, if you had started your timer at the moment you threw the object upwards from the ground you would find that the math worked out to have both t solutions positive, as you would expect since it takes a little bit of time to reach s on the trip up, so the value should be positive, and it takes even more time to reach s on the trip down so you would expect it to be positive too.

The problem with the first situation is that the model is only valid from the time you dropped the object. Previous to that, the model thinks it was acting under the same laws of motion. But it wasn't, you were holding it or whatever. That it why the negative value is not valid in this instance, for negative t, that is the wrong equation.

I'll print this and take a close look at it. Thanks! By the way, um how old are you sir?