Confusing puzzle - clever solution

One of the runners has constant velocity, $v=v_0$ and $s=v_0 t$ . The other runner has an initial velocity and an acceleration, $v'=v_0'+at$ and $s'=v_0't+\frac{1}{2}at^2$ . At time $t_1$ they have the same velocity, $t_1=\frac{v_0-v_0'}{a}$ . At time $t_2$ they are in the same position, $v_0 t_2=v_0't_2+\frac{1}{2}at_2^2$ or

$(v_0-v_0')t_2=\frac{1}{2}at_2^2$ .

Assuming $t_2=\gamma t_1$ we get

$\frac{\gamma(v_0-v_0')^2}{a}=\frac{1}{2}a \gamma^2 \frac{(v_0-v_0')^2}{a^2}$

The solution is $\gamma=2$

We can solve the problem via symmetry:

Suppose the race finishes; we are caught up and you have a higher velocity than I do. Now we race backwards: you start out faster than I do but constantly decelerate. After t seconds, we will have the same speed. In the original race, it took t seconds for us to have the same speed since we started racing, but in the backwards race, it took t seconds for us to have the same speed from when you catch up with me. Thus, it takes a total of 2t seconds for you to catch up with me.

I also devised another solution using areas under the two velocity graphs.

Lets call the runner that starts with instantaneous speed TheFlash , and lets call the other runner TheOtherGuy .

What does it mean for TheOtherGuy to have caught up with TheFlash ? It means they will have moved the same distance from the starting line.

How can we calculate the distance something moves from the starting line?

Well, for TheFlash its easy, we get his velocity and multiply it by the time he's been running for. $V_{Flash} * Time = X$

How about for TheOtherGuy ?

Well, TheOtherGuy doesn't have a constant velocity, so we can't just get his velocity and multiply it by the time he's been running for... If he starts out at a velocity of 0, and after a hundred seconds he is suddenly moving at one meter a second, he can't have gone $1 meter/sec * 100 seconds = 100 meters$ , it wouldn't make any sense.

To calculate the distance something with a constant acceleration moves in a certain amount of time, we need his average velocity over that time period. If he starts out with a velocity of 0, then his average velocity will be the final velocity divided by 2.

So, now we know that for TheOtherGuy , we need to get his final velocity, divide it by 2 to get his average velocity, and multiply it by the time he's been running for to get his distance moved. $V_f/2 * t = X$ Now we need to calculate his final velocity.

We know that after a time t, TheOtherGuy will have a velocity equal to $V_{TheFlash}$ . That means his average velocity will be half of that, and he will only have travelled half as far as TheFlash . We need his final velocity to be twice that of TheFlash in order for his average velocity to be the same as TheFlash's velocity. $\frac{2V{flash}}{2} = V{flash}$ .

And since they will have moved for the same time, that means they will have moved the same distance.

Since TheOtherGuy had a constant acceleration, if he had TheFlash's velocity at time $t$ , he will have twice TheFlash's velocity at time $2t$ . Therefor, at time 2t they will have moved the same distance.

So the answer is $2t$

Hooray!!!

(More) To get the distance the flash moved if we knew his acceleration: $\frac{at}{2}$ $*t =∆X$ , AKA $\frac{at^{2}}{2}$ $=∆X$ Think of why.

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Thank you Laszlo for ur comment! Ok, as Laszlo pointed out, we don't know if TheOtherGuy started out with zero velocity, only that he started slower. So what IF TheOtherGuy doesn't start with zero velocity?

If we have constant acceleration, average velocity over a time interval would be the initial velocity + the final velocity divided by 2. $\frac{V_i + V_f}{2}$ (Now you can see why if we start at zero velocity, the average velocity is simply the final velocity divided by 2)

So, lets say TheOtherGuy starts out with a velocity $V_0$ . TheFlash is moving at some velocity $V_{Flash}$ . So the difference between their velocities is $V_{Flash} - V_0$ .

Now, it takes TheOtherGuy some time $t$ to reach TheFlash's velocity. I don't know how much he will have travelled in that time relative to the Flash...so that part doesn't hold true anymore. However, as you will see, the concept of average velocity multiplied by time to get the distance still holds

I do know that since TheOtherGuy's acceleration is constant, if we wait the same amount of time $t$ again, the difference between his final velocity, $V_f$ , and TheFlash's velocity, $V_{flash}$ , will once again be what it was at the beginning $V_{f} - V_{flash} = V_{flash} - V_0.$

(For example, suppose TheOtherGuy starts out at 7 m/s, and TheFlash is moving at 10m/s. After a time t, TheOtherGuy will be moving at 10m/s. He sped up by 3m/s. If we wait that time t again, TheOtherGuy will be moving at 13m/s, since his acceleration was constant and he will speed up the same amount for the same amount of time.)

We can represent the other guy's initial velocity as $V_{flash} - n$ , where n is some number, and his final velocity (after 2t) as $V_{flash} + n$ , where n is once again that same number.

Therefor, if we add TheOtherGuy's final velocity after time 2t and that initial velocity and divide by 2, we will once again get an average velocity equal to the constant velocity of TheFlash. $\frac{V_{flash} - n + V_{flash} + n}{2} = V_{flash}$ . (or to stick with the same example, (7 + 13) / 2 = 10 )

An average velocity equal to that of TheFlash multiplied by the same time TheFlash has been moving for means, once again, the same distance travelled.

Hooray again! Thanks Laszlo!

Let $v$ be the velocity of "I" runner, and $a$ be the acceleration of the "you" runner, and $t_1, t_2$ be the times the velocities match and positions match. Then:

$v = a t_1$

$v t_2 = \frac{1}{2} a t_2^2$

From this we have $2 t_1 = t_2$