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Let's say an there are two objects or vehicles, both of them travel towards the same direction with constant velocities, the other object being slower than the other. How do you calculate at which point the faster object passes the slower one?

Say the slower object starts it's movement at 50 meters and the faster one starts at 0 meters. If they both keep moving forward with their respective velocities, when does the faster object pass the slower one?

For example, if the slower object has v = 25km/h and the faster one has v = 50km/h. They both start moving at the same time, though the slower one starts 50 meters ahead. When do they pass each other?

I can't think of anything but backbreakingly manually taking advantage of the relativity of their speeds, shouldn't there be some kind of trick to this?

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Two tricks you can use.

First - just solve the equations of motion. You can write down the position as a function of time:

$$x(t) = x(0) + v\cdot t$$

where $x(t)$ is the position at time $t$, and $v$ is the velocity. Two equations, solve for $t$.

Simpler still - look at the difference in velocity. If one goes at 25 km/h and the other goes at 50 km/h, the faster one is catching up on the slower one at a speed of (50-25)=25 km/h. So whatever the gap between them at the start, that's the gap that he is closing at that speed.

Then the time taken to close the gap is (initial gap) / (speed of closing the gap), and once you have the time, you can calculate the distance traveled because you have the speed.

Think you can take it from here? I hope you see it doesn't have to be "back breaking".

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As you propose let's find a "trick" and take advantage of the relative speed of the vehicles.

Let's find out when they meet first. Let's imagine we are sitting in the slower vehicle. We say we are "in a frame of reference attached to the slower vehicle". From our perspective the faster vehicle is moving towards us at 50-25 = 25km/h. Still from our perspective the slower vehicle is not moving as we are sitting in it. The faster vehicle has 50m to go to pass us. Time = Position / Speed so the faster vehicle shall meet us in t = 50m / 50000km/h = 0.001 hours = 3.6 seconds.

Let's get back to the perspective of an external observer now. Our frame of reference is now attached to the Earth again. Position = Velocity x Time so in 3.6 seconds the slower object has moved 0.001h x 25km/h = 50m and the faster object has moved 0.001h x 50km/h = 100m.

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If you can write the answer without the equation written down, you can input the values into a graphing calculator. (input it in the Y= table) For example, Y1= 50(X) <- Object that started at 0m but is faster. Y2=25(X)+50 <- Object that started ahead, but is slower. X= seconds, of course. Now, just go into the table (2nd+Graph) And find where the numbers line up equally.

If you input this into a graphing calculator, you get 100m.

It may not be perfect, but it works for me.

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