Shreya said:
Awesome! Does that mean that the equations we get in 3 & 4 doesn't describe the range of ##u##s and the corresponding ##t##s we can get, but only a specific situation.
Another Silly Question - then why do we still get a function (not a number)? And doesn't the lower values of u from 3 & 4 satisfy the conditions?
Essentially you do get a number (really a quantity, as we have units involved). There's a difference between a fixed, as yet uncalculated quantity and a variable. To take a simple example.
We start with the simple equation relating displacement, constant velocity and time:
$$s = vt$$In general, all of these quantities may be variables. Note that there is a subtle difference between ##v##, which is an unknown, and ##s## and ##t## which are continuous variables.
In a single experiment, ##v## would be some fixed quantity throughout the experiment, while ##t## and ##s## vary.
If we consider
varying ##v##, then we are considering many different experiments. This is the technique you tried to use for this problem. This is a valid approach, but you must be aware of what you are doing.
If we want to hit a target at a distance of ##10 \ m## then the speed and time are related by:
$$vT = 10 \ $$Which gives ##v## as a function of ##T## and vice versa. We can plot a graph of ##v## against ##T##. Note that I've been careful to use ##T## here to indicate that we have a fixed but unknown time ##T##. In any experiment we have a single ##v## and a single value for ##T##.
This ##T## is not the same as the continuous variable ##t## that represents the passage of time.
In this case, there is no minimum of maximum, but we can still draw a graph of the function ##v(T) = \frac{10 \ m}{T}## or ##T(v) = \frac{10 \ m}{v}##.
Note that this is a function that represents many different experiments, with all the possible values of ##v##. In any given experiment, ##v## and ##T## are fixed (even if unknown) quantities.
Finally, if we assume that ## v = 5 \ m/s##, say, then we no longer have a function and ##T## must be ##2 \ s##. In this simple case, as soon as we choose either ##v## or ##T## then the function collapses to a single point on the graph. This is what happened above when you chose ##v = \ 25 m\s## as the final speed. You had then specified enough data to collapse the function relating ##u## and ##T## to a single point on the graph (##u = 75 \ m/s## and ##T = 5 \ s##). Even if you still had to calculate these numbers.
In particular, you no longer had a function ##u(T)## to do calculus on.
