Mathematical Analysis Of V^2 = Vo^2 + 2a(X-Xo)

[babaliaris]

I found the formula but I can't understand how to interpret the 3 cases.

Case 3 make sense. d > 0 if a > 0 because Vo^2 > 0 and Δx > 0 by default (can't have negative length). In this case t2 is negative because the numerator is a sum of two negative values and the denominator a > 0 so we reject it because we can't have negative time. This is how I ended up using t1 from case 3 to substitute into the V = Vo + at

But what the other two cases mean? What it mean d < 0 or d = 0 to the original equation?

 

[babaliaris]

I have already tried to solve for t in the equation V = Vo + at and substitute it into the Δx equation
The reason I did the above hard way (solving a second degree equation) is just for fun

 

[Merlin3189]

I'm struggling to follow your algebra, but I understand you got a quadratic for t.
Now you can use the formula, as you have done, but there is no need to consider 3 cases.
Just write down the solution as you do for case 3. The solution should have the same form for all three cases.

I think you are wrong to say that you can't have negative time or negative distance. Both of these occur in the solution of problems like this.
These formulae apply to problems of constant acceleration. That means that the equations assume the body is accelerating, will be accelerating and always was accelerating at this constant rate. A solution where t = -5 for eg. simply means the time 5 sec before t₀ and tells you what the position and speed would have been, IF the constant acceleration assumption had been true.

When you apply these formulae to situations where the acceleration is not constant over all time, eg. the object started moving at time t₀, then the answer with negative time is not relavent to your situation. Someone observing the moving object at some later time would not know when it started, so would see the negative time solution as equally valid. They would be unable to distinguish between an object that started at t₀ and one that followed the same trajectory and had been moving for ever.

Negative distance is even more simple. Just like negative velocity or negative acceleration, it just means the opposite direction.
Eg. If you throw a ball from a tall building, at some point it may be above or below the starting point.
Similarly, an object which is above me when standing on the ground, may be below someone standing on a tall bbuilding. If we apply the same equations to the same object, but measure from our own position, we may get different numbers for the same position.

Don't ignore the two solutions of a quadratic. Look at both, but be aware some solutions may be outside the range of possible real answers.

Anyhow, you don't need to get into quadratics here. You don't require t in your solution, so just eliminate t from your two starting equations. It's 4 to 6 lines of simple algebra. Absolutely no quadratics. (I see you just explained this!)

 

[babaliaris]

Thank you! Your answer gave me some more knowledge about what mathematics are actually telling me. So actually negative times and lengths in these equations do not actually make sense since we have defined them with starting time at 0s and starting position at 0m.

But if I'm not mistaken this is what you are talking about:
$$ a = a_{avg} = \frac{v-v_0}{t-t_0} \Leftrightarrow a(t-t_0) = v-v_0 \Leftrightarrow v = v_0 + (t-t_0)$$
$$v_{avg} = \frac{x-x_0}{t-t_0} \Leftrightarrow x-x_0 = v_{avg}(t-t_0)$$
Because $$v_{avg} = \frac{v_0+v}{2}$$
We get $$x-x_0 = v_0(t-t_0) + \frac{1}{2}a(t-t_0)^2$$

Now in these equations, does a negative length or time make sense?

 

[PeroK]

Thank you! Your answer gave me some more knowledge about what mathematics are actually telling me. So actually negative times and lengths in these equations do not actually make sense since we have defined them with starting time at 0s and starting position at 0m.

But if I'm not mistaken this is what you are talking about:
$$ a = a_{avg} = \frac{v-v_0}{t-t_0} \Leftrightarrow a(t-t_0) = v-v_0 \Leftrightarrow v = v_0 + (t-t_0)$$
$$v_{avg} = \frac{x-x_0}{t-t_0} \Leftrightarrow x-x_0 = v_{avg}(t-t_0)$$
Because $$v_{avg} = \frac{v_0+v}{2}$$
We get $$x-x_0 = v_0(t-t_0) + \frac{1}{2}a(t-t_0)^2$$

Now in these equations, does a negative length or time make sense?

You can't have a negative length, by definition, but you can have a negative displacement. In fact, in any case where you have a displacement along the x-axis, that displacement is positive or negative depending on the orientation of your x-axis.

If by a negative time you mean $t < t_0$, then that makes sense also. The kinematic equations can not only be used to calculate where something will be at a future time but also to calculate where something was at an earlier time. Sometimes $t_0$ may be in the middle of an experiment and you could calculate the position of an object at times $t = t_0 \pm1s$, say.

Note also that a large part of cosmology is figuring out what happened in the past! All the data is from the present (last hundred years) and the cosmological models need to be valid if you run time backwards towards the big bang.

When you first meet $x_0, t_0$ you may think of them as the starting point. But, more generally, you can think of them as some reference point that may be in the middle of an experiment.

 

[babaliaris]

Awesome! I got it now!