# How long to catch up to a stream started $1$ hour ago at $1.5$x speed?

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?

• 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. – saulspatz May 29 '20 at 20:33
• 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. – Aravindh Krishnamoorthy May 30 '20 at 14:01
• 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). – Glen O May 31 '20 at 7:22

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.

• So the final equation for my usage is $t_{total} = \frac{t_{start}}{s_{playback}-1}$. That's easier than I was expecting. – Kevin Fee May 30 '20 at 0:56

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. $$t=\frac23(1+t)\implies t=2$$

• 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. – Bernhard Barker May 30 '20 at 22:21
• @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. – saulspatz May 31 '20 at 0:39

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 $$t = 2 hr$$.

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

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

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?)

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