I don't think we did well by this student. That is only partially his fault: he did not tell us upfront that what he did not understand was the explanation he had already received for this problem, which put us majorly off the track, and he made an unwarranted assumption, which suggested that he did not understand the problem. But the explanation that he had received is a disgrace. My mind would have been scrambled too had I listened to all 50 minutes of that video.
Tkhunny is correct; the problem as posed only makes sense in the context of an optimization problem. Having looked at the cited video, however, its audience obviously is innocent of optimization. That we are trying to find a formula for the minimum of a variable is a hidden assumption, and those can only confuse students, particularly good ones. I am going to assume that the underlying optimization problem was solved correctly (the solution looks plausible, but I did not prove it was correct) and address the problem actually presented to the viewers of the video, which clearly contemplates maximum acceleration and deceleration.
And Subhotosh Khan is correct: there is no reason to assume that a and d are equal. But once you see the hash made of notation in the video, the student's assumption is at least excusable..
I have looked at the relevant part of the video that presents and explains this problem. I know how to solve the problem and still found the video's explanation rapid and opaque.
The video uses the same letter to stand for a variable in some cases and for a constant in others. That is, s is a variable in some formulas, but a constant in this problem, and is not used in the same sense as in the formulas. The formulas use u to mean the initial velocity when acceleration changes, but acceleration changes three times in this problem, and there is no variable u as this problem is presented. t sometimes mean the time spent at constant acceleration and sometimes mean a sum of times at different accelerations. The notation is decipherable, but it must be a nightmare for students.
In this problem, we have a period of constant acceleration immediately followed by a period of constant deceleration. The object of the exercise is to derive BY HAND a formula for the time when velocity is changing, a variable that I call y.
I am going to follow two sensible things in the video. One is to use subscripts to distinguish between variables in the acceleration and deceleration phases. The other is to make consistent the names of unknowns in the problem with variables in the general formulas that are provided.
We have, in the completely general case, where a is acceleration, u is initial velocity, v is final velocity, s is distance travelled under constant acceleration, t is time spent at constant acceleration, x is the total distance traversed while velocity is changing, and y is the total spent while velocity is changing, this system:
[MATH]v_1 = u_1 + a_1 t_1.[/MATH]
[MATH]s_1 = \dfrac{u_1 + v_1}{2} * t_1 \implies 2s_1 = (u_1 + v_1)t_1.[/MATH]
[MATH]v_2 = u_2 - a_2 t_2.[/MATH]
[MATH]s_2 = \dfrac{u_2 + v_2}{2} * t_2 \implies 2s_2 (u_2 + v_2)t_2.[/MATH]
[MATH]v_1 = u_2.[/MATH]
[MATH]x = s_1 + s_2 \implies 2x = 2s_1 + 2s_2.[/MATH]
[MATH]y = t_1 + t_2.[/MATH]
In other words, we have 12 unknowns and only 7 equations in the general case. This of course is not solvable at all. But, in our specific problem, a1, a2, and x are supposed to be positive constants. I shall use Greek letters for what are deemed constants. Furthermore, u1 and v2 are given as zero. That takes us down to seven unknowns in seven equations, which means that it may be ( and in fact is) feasible to solve the system. Moreover, because v1 = u2, we can easily eliminate one of those unknowns, thus reducing the problem to six equations in six unknowns. Solving by hand six simultaneous equations, two of which are not linear, is a pain in the neck. But it is not conceptually difficult. Students can follow an orderly, systematic attack. The system reduces to this.
[MATH]v_1 = \alpha t_1.[/MATH]
[MATH]2s_1 = v_1t_1.[/MATH]
[MATH]0 = v_1 - \delta t_2.[/MATH]
[MATH]2s_2 = v_1t_2.[/MATH]
[MATH]2 \sigma = 2s_1 + 2s_2.[/MATH]
[MATH]y = t_1 + t_2.[/MATH]
But we can easily substitute for s1 and s2 and reduce to four equations in four unknowns.
[MATH]v_1 = \alpha t_1.[/MATH]
[MATH]0 = v_1 - \delta t_2.[/MATH]
[MATH]2\sigma = v_1t_1 + v_1t_2 = v_1(t_1 + t_2).[/MATH]
[MATH]y = t_1 + t_2.[/MATH]
And it is easy to eliminate v1 to reduce to three equations in three unknowns.
[MATH]0 = \alpha t_1 - \delta t_2.[/MATH]
[MATH]2 \sigma = \alpha t_1(t_1 + t_2).[/MATH]
[MATH]y = t_1 + t_2.[/MATH]
The first of those equations
[MATH]\implies t_2 = \dfrac{\alpha t_1}{\delta} \implies t_1 + t_2 = t_1 +\dfrac{\alpha t_1}{\delta} = t_1 * \dfrac{\alpha + \delta}{\delta}.[/MATH]
This permits a simplification to two equations in two unknowns.
[MATH]y = t_1 * \dfrac{\alpha + \delta}{\delta}.[/MATH]
[MATH]2 \sigma = \alpha t_1 * \left ( t_1 * \dfrac{\alpha + \delta}{\delta} \right) = t_1^2 * \dfrac{\alpha (\alpha + \delta)}{\delta} \implies[/MATH]
[MATH]t_1^2 = \dfrac{2 \delta \sigma}{\alpha(\alpha + \delta)} \implies t_1 = \sqrt{\dfrac{2 \delta \sigma}{\alpha(\alpha + \delta)}} \implies[/MATH]
[MATH]y = \sqrt{\dfrac{2 \delta \sigma}{\alpha(\alpha + \delta)}} * \dfrac{\alpha + \delta}{\delta} = \sqrt{\dfrac{2 \delta \sigma}{\alpha(\alpha + \delta)}} * \sqrt{\left ( \dfrac{\alpha + \delta}{\delta} \right )^2} \implies[/MATH]
[MATH]y = \sqrt{\dfrac{2 \delta \sigma}{\alpha(\alpha + \delta)} * \left ( \dfrac{\alpha + \delta}{\delta} \right )^2} = \sqrt{\dfrac{2 \delta \sigma}{\alpha (\alpha + \delta)} * \dfrac{(\alpha + \delta)^2}{\delta^2}}\implies[/MATH]
[MATH]y = \sqrt{\dfrac{2 \sigma (\alpha + \delta)}{\alpha \delta}}. \text { Q.E.D.}[/MATH]
What a time-wasting, mind-numbing exercise, but it does not involve anything conceptually difficult, just painstaking care and basic algebra.
There is no way that a student is going to follow the derivation of that result in ten minutes of babble about things like lateral thinking.
