Two points move in the same straight line starting at the same moment from the same point in it. The first moves with constant velocity 'u' and the second with constant acceleration 'f'. During the time elapses before the second catches the first greatest distance between the particle is?

The answer is u^2/2f but i don't know why?

Question

Two points move in the same straight line starting at the same moment from the same point in it. The first moves with constant velocity 'u' and the second with constant acceleration 'f'. During the time elapses before the second catches the first greatest distance between the particle is?

The answer is u^2/2f but i don't know why?

theeric · Answer

Okay! Well, if the starting time is $t=0$ and starting distance $x=0$, which is very convenient, then we want to find what $t$ is when the two points have the same $x$ value.

Each point's distance traveled depends on the time since the start. The accelerating point depends on its acceleration, $f$, and the constant speed point depends on its velocity, $u$. We'll give them separate distance variables. I'll use $x_a$ for the accelerating point and $x_c$ for the constant velocity point.

Remember, both depend on time. Given only the variable time, you can find the $x$'s by using the constant $u$ and $f$.

$x_a = x_0 + u_0\ t + \frac{1}{2}f\ t^2=\frac{1}{2}f\ t^2$
$x_c=x_0+u\ t=u\ t$

So, to sum that up,
$x_a=\frac{1}{2}f\ t^2$   and   $x_c=u\ t$.

At what $t$ are they equal? Well, to find out, set them equal and find what $t$ must be!
$x_a=\frac{1}{2}f\ t^2=x_c=u\ t$
So, $\frac{1}{2}f\ t^2=u\ t$
Use some algebra, now...
Divide both sides by $\frac{1}{2}f\ t$ to get $\dfrac{\frac{1}{2}f\ t^2}{\frac{1}{2}f\ t}=\dfrac{u\ t}{\frac{1}{2}f\ t}$
Look for cancellations, now. $\dfrac{\cancel{\frac{1}{2}}\cancel{f}\ t^\cancel{2}}{\cancel {\frac{1}{2}}\cancel f\ \cancel t}=\dfrac{{u\ \cancel t}}{\frac{1}{2}f\ \cancel t}$
You are left with $t=\dfrac{u}{\frac{1}{2}f}$, which is $t=\dfrac{2\ u}{f}$

Please feel free to ask for clarification if necessary!

anonymous · Answer

Thanks for responding!

You said that we need to find the t when both the x or the distance traveled by them is same. 
Why do we need to find that?

theeric · Answer

Whoa, I read the question wrong, so what I found was the time at which the two points pass each other.

So there are two ways to approach the actual problem. One way is to create an equation for the distance between the two points find the first maximum.

But the physics way is easier, I think. We'll see. See, the particle with constant-velocity is going faster at first. It keeps on covering more ground than the accelerating particle... until when?

When the accelerating particle gets to the velocity of the constant-velocity particle, they are instantaneously covering the same ground. Then the accelerating particle will accelerate, meaning it will get faster than the constant-velocity particle. Then the accelerating particle will start catching up. Then the distance between the two particles will decrease. So we want to see the distance between the two particles when the accelerating particle reaches the constant-velocity particle.

theeric · Answer

I'll use $x_a$ and $x_c$ again.

We need to know how long it takes for the accelerating particle to accelerate to the velocity $u$. That is because we'll be able to then have a time so we can know how far the constant-velocity particle got in that time, and we can subtract how far the accelerating particle got to see the gap between them.|dw:1376862036517:dw|

theeric · Answer

$\huge\sf \color{blue}{Finding\ the\ time\ at\ which}$
$\huge\sf\color{blue}{the\ accelerating\ particle}$
$\huge\sf \color{blue}{attains\ a\ speed\ of\ \boldsymbol u.}$
$a=\dfrac{v_f-v_i}{t_f-t_i}=\dfrac{v_f}{t_f}$$\implies t_f=\dfrac{v_f}{a}$
For our acceleration and velocity, we have $f$ and $u$, respectively. So
$t_f=\dfrac{u}{f}$

How far did each particle go?

theeric · Answer

$\huge\sf \color{orange}{Finding\ \boldsymbol{x_a}\ at\ \boldsymbol {t_f}.}$
$d_f=d_i+v_i\ t+\frac{1}{2} a\ t^2$
That is,
$d_f=\frac{1}{2} a\ t^2\implies x_c=\frac{1}{2}f\ t_f=\frac{1}{2}f\ \dfrac{u}{f}=\frac{1}{2}u$
In summary, $x_c=\dfrac{u}{2}$.

theeric · Answer

$\huge\sf \color{green}{Finding\ \boldsymbol{x_c}\ at\ \boldsymbol {t_f}.}$
$v=\dfrac{d_f-d_i}{t_f-t_i}\quad\leftarrow$definition of (average) velocity
For us, the initial values are $0$, so
$v=\dfrac{d_f}{t_f}\implies d_f=v\ \ t_f$

With our values, that is
$x_c=u\ \dfrac{u}{f}=\dfrac{u^2}{f}$

In summary, $x_c=\dfrac{u^2}{f}$.

theeric · Answer

$\huge\sf \color{red}{Finding\ \boldsymbol{x_c-x_a}.}$
$x_c=\dfrac{u^2}{f}$

$x_a=\dfrac{u}{2}$   I incorrectly labeled it $x_c$ in its section, sorry.

$x_c-x_a=\dfrac{u^2}{f}-\dfrac{u}{2}=u\left(\dfrac{u}{f}-\dfrac{1}{2}\right)$

$\color{red}{\text{Where did I go wrong? :(}}$
$$\it\Huge\color{blue}{ ?...}$$

anonymous · Answer

Hey thanks!

theeric · Answer

But my answer didn't match yours...

anonymous · Answer

I will certainly write that as i have figured out that part but here's a doubt. 
The question asks the maximum distance between the two points before the second catches with the first. 
Now i am a bit confused,does it mean the distance between them when the second point's (the one with the acceleration 'f')  velocity is 'just' less then the first point or does it mean when the subtraction of the total distance traveled by each point from origin is maximum.

theeric · Answer

I'm thinking it wants to know the distance between particles. From you question, I quote, "greatest distance between the particle is?"

And think about this - when the accelerating particle reaches $u$, if it $\sf stopped$ acceleration, they would both have speed $u$. If they continued like that forever, the distance between them would not change. But it will instantly be smaller if the particle started accelerating again. So that distance between them would shrink. The distance between them when both have the same speed is going to be the largest distance between them before they pass each other.    We shouldn't look for the distance between them just before the accelerating particle achieves $u$, because that would neglect the distance gap gained from "just before $u$" to "$u$."    And, as I said, the gap will really be largest when they both travel at $u$.

Is that alright?

anonymous · Answer

hey i got that when the second point's(accelerating point) velocity crosses the first it is only going to get closer but what about the other case when its velocity is less then the first? How do i rule this out?
Can't the max distance be when its velocity is still less then the first point?

amistre64 · Answer

think of a speeding car, that passes by a parked police car; at t=0

Cv(0) = u;  Pv(0) = ft

Cd = ut, Pd = f t^2/2

the distance between them is defined as Cd - Pd, which creates an inverted parabola when graphed

ut - f t^2/2,  the maximum of this distance would be located at the vertex ... when t = u/2f

u(u/2f) - f (u/2f)^2/2

the speeding car is at u^2/2f ; while the police car is at u^2/8f

anonymous · Answer

if we take a=-f/2 and b=u then which equation i should use to solve for t?
y=ax^2 +bx +c or x=ay^2 + by + c

amistre64 · Answer

we are given a constant speed;  v(t) = u
we are given a constant acceleration:  a(t) = f

constant acceleration gives us linear velocity:

a(t) = f , therefore v'(t) = ft, adjusted for v(0)=0

distance covered equates to the area under the "curve" of each graph:

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