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Anama Skout
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Consider the famous kinematics equations $$x_f=x_i+v_i\cdot t+\frac12at^2$$ My question is: does this equation work for every point A & B along the parabola formed when dropping an object?
Thanks for your answer. But I have something different in mind. Suppose that the point A has ##x_{A,0}## and ##v_{A,i}##, then the equation will have the form $$x_A(t)=x_{A,i}+v_{A,i}\cdot t+\frac12at^2$$ and for a point B which has ##x_{B,0}## and ##v_{B,i}## then the equation will have the form $$x_B(t)=x_{B,i}+v_{B,i}\cdot t+\frac12at^2$$ Will this yield the same values in that for instance if we use them to determine what will happen after ##t=5## as long as we know what ##x_{A,0}##, ##v_{A,i}##, ##x_{B,0}##, and ##v_{B,i}## are (and the time difference between the two)?DaleSpam said:Yes. Each point is represented by a different t. To make that clear, I would write it as:
$$x(t)=x_0+v_0 t+\frac{1}{2}a t^2$$
Why don't you try it and see for yourself?Anama Skout said:Will this yield the same values in that for instance if we use them to determine what will happen after ##t=5## as long as we know what ##x_{A,0}##, ##v_{A,i}##, ##x_{B,0}##, and ##v_{B,i}## are (and the time difference between the two)?
I tried it with three points, always worked well (except the times when I made some mistakes -- of course), however I'm more interested into a rigorous proof.Doc Al said:Why don't you try it and see for yourself?
Should be doable with a bit of algebra. Go for it!Anama Skout said:however I'm more interested into a rigorous proof.
The given form of the equation only works for ##x_0## and ##v_0## defined at ##t=0##. At ##t=0## the object is only at one point. If you use the same formula for a point which is on the parabola at a different time then you are going to get a different parabola which has that point at ##t=0##.Anama Skout said:Thanks for your answer. But I have something different in mind. Suppose that the point A has ##x_{A,0}## and ##v_{A,i}##, then the equation will have the form $$x_A(t)=x_{A,i}+v_{A,i}\cdot t+\frac12at^2$$ and for a point B which has ##x_{B,0}## and ##v_{B,i}## then the equation will have the form $$x_B(t)=x_{B,i}+v_{B,i}\cdot t+\frac12at^2$$ Will this yield the same values in that for instance if we use them to determine what will happen after ##t=5## as long as we know what ##x_{A,0}##, ##v_{A,i}##, ##x_{B,0}##, and ##v_{B,i}## are (and the time difference between the two)?
DaleSpam said:The given form of the equation only works for ##x_0## and ##v_0## defined at ##t=0##. At ##t=0## the object is only at one point. If you use the same formula for a point which is on the parabola at a different time then you are going to get a different parabola which has that point at ##t=0##.
In other words, unless ##A=B## the equations that you have written represent two distinct parabolas. I.e. the first equation passes through ##A## at ##t=0## and the second equation passes through ##B## at ##t=0##, they are not the same parabola even if both parabolas pass through ##A## and ##B##.
Also, I am not sure if you are thinking about parabolas through space or parabolas in spacetime. The equation you have written is for 1D motion, so it is only a parabola in spacetime.
I'm thinking of parabolas in space parametrized w.r.t time. (like the one I drawed)DaleSpam said:So are you thinking of parabolas in space parametrized by time or are you thinking of parabolas in spacetime?
I only treated the system w.r.t to a coordinate system in which the ##x##-axis is vertical.Doc Al said:Why don't you treat each axis separately? I would think that would be simpler.
Ah, so you are only talking about the vertical axis (as I first thought) despite the mention of parabolas. OK.Anama Skout said:I only treated the system w.r.t to a coordinate system in which the ##x##-axis is vertical.
I am not quite sure why you want to put yourself through this, but OK. So a parabola in space can be defined by the parametric equations:Anama Skout said:I'm thinking of parabolas in space parametrized w.r.t time. (like the one I drawed)
How would this equation work for parabola? This is a straight line equation isn't it?DaleSpam said:Yes. Each point is represented by a different t. To make that clear, I would write it as:
$$x(t)=x_0+v_0 t+\frac{1}{2}a t^2$$
The motion is along a straight line. But expressed as a function of time, the equation describes a parabola.Hamza Abbasi said:How would this equation work for parabola? This is a straight line equation isn't it?
The confusion here is that you have defined too many variables. Point i or 0, is a chosen initial point on the trajectory. A and B are two other points on the same trajectory. So there is no separate initial position for A and a separate one for B.Anama Skout said:Thanks for your answer. But I have something different in mind. Suppose that the point A has ##x_{A,0}## and ##v_{A,i}##, then the equation will have the form $$x_A(t)=x_{A,i}+v_{A,i}\cdot t+\frac12at^2$$ and for a point B which has ##x_{B,0}## and ##v_{B,i}## then the equation will have the form $$x_B(t)=x_{B,i}+v_{B,i}\cdot t+\frac12at^2$$ Will this yield the same values in that for instance if we use them to determine what will happen after ##t=5## as long as we know what ##x_{A,0}##, ##v_{A,i}##, ##x_{B,0}##, and ##v_{B,i}## are (and the time difference between the two)?
Equation is of straight line but it explains parabola?? How come?Doc Al said:The motion is along a straight line. But expressed as a function of time, the equation describes a parabola.
If you plot x as a function of time, it is not a straight line.Hamza Abbasi said:Equation is of straight line but it explains parabola?? How come?
It is a straight line in space. It is a parabola in spacetime.Hamza Abbasi said:How would this equation work for parabola? This is a straight line equation isn't it?
Wow! I didn't knew that , can you please post some pictorial representation?DaleSpam said:It is a straight line in space. It is a parabola in spacetime.
You have the function. Plot it for yourself!Hamza Abbasi said:Wow! I didn't knew that , can you please post some pictorial representation?
If I understand you correctly, ##x_{B,i}## and ##v_{B,i}## are the position and velocity of the particle some time (say ##\Delta t##) after it passes through ##x_{A,i}##. So all you need to show is that the position given by the first equation at time ##t + \Delta t## is the same as the position given by the second equation at time ##t##. That's why I said it was just a bit of algebra.Anama Skout said:Thanks for your answer. But I have something different in mind. Suppose that the point A has ##x_{A,0}## and ##v_{A,i}##, then the equation will have the form $$x_A(t)=x_{A,i}+v_{A,i}\cdot t+\frac12at^2$$ and for a point B which has ##x_{B,0}## and ##v_{B,i}## then the equation will have the form $$x_B(t)=x_{B,i}+v_{B,i}\cdot t+\frac12at^2$$ Will this yield the same values in that for instance if we use them to determine what will happen after ##t=5## as long as we know what ##x_{A,0}##, ##v_{A,i}##, ##x_{B,0}##, and ##v_{B,i}## are (and the time difference between the two)?
Oups, another mistake, I wanted to say that we would get the same result between these points if we translate one to another. I'll edit the earlier post.nasu said:Well, why use x when all your labels in the picture are y? But this is a minor point.
You did not make any mistake.
The position is -22.5 m in respect to A but -42.9 m in respect to B. Did you expect to get the same value? You are measuring (same) position from different origins. It will be unlikely to get the same value. If you look at the figure you can see that the distance from B to D is larger than from A to D.(or than C to D).
In this case I translated A to B, the difference between the x-coordinate of the two is only 2.04 m, and doesn't help to get 22.5 m out of the 43 m, why?nasu said:I don't know what you mean by this "translate one to another". By changing the origin, the values of the coordinate will change and there is no mistake.
The positions will change but speed and acceleration won't change if you translate the origin to a different point along the same line.
Oh, sorry again for the mistake. Thanks for that, now it become all clear!nasu said:No, the difference is 20.4 m.
A has 0 coordinate and B has 20.4 m. You are mixing coordinate and time.
Because the speed is the derivative of the position in respect to time. Adding a constant to the coordinate (this is what a translation means) does not change the derivative.
I get this $$\overline{v}_1=\frac{\Delta x}{\Delta t}=\frac{(x_B-x_D)-(x_B-(x_D-1))}{t_{B;D}-t_{B;D'}}=-\frac1{t_{B;D}-t_{B;D'}},$$ where ##t_{B,D}## means the time the particle is at D relative to B, and D' stands for the point when the stone moves 1m below D. And with respect to C $$\overline{v}_2=\frac{\delta x}{\delta t}=\frac{(x_C-x_D)-(x_C-(x_D-1))}{t_{C;D}-t_{C;D'}}=-\frac1{t_{C;D}-t_{C;D'}},$$ so to show that these two things are equal one would require that $$t_{B;D}-t_{B;D'}=t_{C;D}-t_{C;D'}.\tag1$$ I'm stuck here...nasu said:Well, look at finite changes. You can think about an average speed, which is Δx/Δt.
Calculate the change in position for two different origin.
For example, if the stone was at D and it moves 1m below D, by how much increases the distance between the stone and C or B?
A freely falling object is an object that is only acted upon by the force of gravity. This means that it is not being pushed or pulled by any other forces, such as air resistance or friction.
A freely falling object accelerates because of the force of gravity. As the object falls, it gains speed due to the constant acceleration caused by the pull of gravity.
The mass of a freely falling object does not affect its acceleration. According to Newton's Second Law of Motion, the acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass. Since the only force acting on a freely falling object is gravity, its mass does not affect its acceleration.
The acceleration of a freely falling object is approximately 9.8 meters per second squared (m/s^2), which is the acceleration due to gravity on Earth. This value may vary slightly depending on the location and altitude.
Yes, a freely falling object can reach a constant speed if there is no longer a net force acting on it. This can happen when the object reaches the ground or when it reaches terminal velocity, the maximum speed an object can reach due to air resistance.