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I opened a stream that started an hour ago. Not wanting to miss anything, I started from the beginning and set it to 1.5x speed. How long will it take for me to catch up?

I know that it will take 40 minutes to watch the hour that I missed ($ \frac {60}{1.5}=40$) but during that time, the stream has generated another 40 minutes that I need to watch. This tells me I need to do some calculus, but it's been a decade since I took that course. Can someone help me come up with an equation?

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  • $\begingroup$ Do you remember geometric series? It takes $\frac23$ as long to watch the next segment, during which it generates $\frac23$ as much additional content, and so on. $\endgroup$
    – saulspatz
    May 29, 2020 at 20:33
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    $\begingroup$ Although this question is very basic, it's interesting because it shows that, like the bird and the train question, there's an easy way and there's a hard way to look at things. The hard way is to realize that every t generates 2t/3 new content and solve as a calculus problem or a series problem. The alternative is to say that you catch up 30 mins of extra content every hour at 1.5x, so you'll catch up an hour's content in 2 hours, which doesn't invoke anything outside of elementary mathematics. $\endgroup$
    – user532446
    May 30, 2020 at 14:01
  • $\begingroup$ I just thought I'd point out that, knowing that you gain 20 minutes per hour, since you started 1 hour behind, it is easy to see that it will take 3 hours to catch up a full hour (three lots of 20 minutes). $\endgroup$
    – Glen O
    May 31, 2020 at 7:22

6 Answers 6

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The easy way to do this doesn't require calculus, or even geometric series. Say it takes $t$ hours to catch up, so you have viewed $1+t$ hours of content at $\frac32$ speed. $$t=\frac23(1+t)\implies t=2$$

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    $\begingroup$ Minor subjective point: I would put the $\frac23$ on the other side of the equation (as $\frac32$). $t$ hours from now, the amount of content you'd have watched is $\frac32t$ and the progress of the stream would be $t+1$ and you want those to be equal. That's much more understandable to me. $\endgroup$ May 30, 2020 at 22:21
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    $\begingroup$ @B Or on the other hand, $t$ is $\frac23$ of the time it would normally take to watch $1+t$ hours of content. Six of one, half a dozen of the other. $\endgroup$
    – saulspatz
    May 31, 2020 at 0:39
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To make this easier to understand, I will solve this as if you are an object travelling at $1.5m/s$, and the stream is an object which started one hour before you travelling at $1m/s$ in the same direction. So you are travelling at $1.5\times$ the speed of the stream.

We know that disance ($d$), speed ($s$), and time ($t$), are related as follows: $$\Delta d=\Delta s\Delta t$$ You are trying to find time, so rearranging for $t$ gives $$\Delta t=\frac{\Delta d}{\Delta s}$$ The initial "distance" between you and the stream is $3600$ metres, based on a speed of $1m/s$ for one hour. So $$\Delta d=3600m$$ The difference between your speed and the stream's speed is $1.5m/s-1m/s=0.5m/s$. So $$\Delta s=0.5m/s$$ Now solving for $\Delta t$: $$\Delta t=\frac{3600m}{0.5m/s}=7200s$$ So, it will take you $7200s$, or $2$ hours, to catch up with the stream.

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    $\begingroup$ So the final equation for my usage is $t_{total} = \frac{t_{start}}{s_{playback}-1}$. That's easier than I was expecting. $\endgroup$
    – Kevin Fee
    May 30, 2020 at 0:56
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You are watching at rate $1.5t$. The stream is playing at a rate of $1 hr + 1t$. The intersection of these two lines occurs when $1.5t = 1 hr + 1t$, or $t = 2 hr$.

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I don't recommend solving this as a geometric series, but it can be solved as one:

The first $60$ minutes is watched in $60/1.5 = 40$ minutes, the next $40$ minutes is watched in $60/1.5^2 = 40/1.5 = 26.66\ldots$ minutes, and so on. Writing this as a sum, we have

$$ T = {60\over1.5} + {60\over1.5^3} + {60\over1.5^3} + \cdots $$

This is often written using the summation symbol:

$$ \sum_{i=1}^\infty {60\over1.5^i} $$

There is a neat identity for infinite series like this,

$$ \sum_{i=\color{orange}0}^\infty ar^i = {a\over1-r} $$

Noting that this series starts with $0$, we add and subtract $60/1.5^0$ from out series:

$$ \color{blue}{-60/1.5^0} + \color{blue}{60/1.5^0} + \sum_{i=1}^\infty {60\over1.5^i} = \color{blue}{-60/1.5^0} + \sum_{i=\color{blue}0}^\infty {60\over1.5^i} = -\color{blue}{60} + \sum_{i=0}^\infty {60\over1.5^i} $$

With $a = 60$ and $r = 1/1.5$, we can use the identity to solve the equation

$$ -60 + \sum_{i=0}^\infty {60\over1.5^i} = -60 + {60\over 1 - 1/1.5} = -60 + 180 = 120 $$

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Here is just a thought (NOT FULL ANSWER) :

Consider two particles A and B on the x-axis. A is at ($60,0$) while B is at origin. Both move towards the positive x-direction ; A with speed $1unit/min$ and B with $1.5unit/min$.

How much time will B take to catch up to A?

(HINT: Have you heard about relative motion?)

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The distance to be covered is the length of video on screen(that is the distance the seeker has moved and is still moving)

Your speed relative to the live-stream is 0.5v where v is the live-stream speed (distance moved by regular seeker on screen divide by time in secs)

Relative distance to be covered by the faster seeker to reach the slower live seeker is $3600*v$

Therefore time required for you to catchup is $3600v/0.5v=7200 secs$ Or two hours

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