Hello Everyone

While solving the above question, we had considered that **after 2 hours John starts walking** and **Franks stops walking** and continued with further procedures whereas

in this question, we had considered that **the boy is walking when the truck crossed him and breaks down after travelling a distance of 1 kilometer.**

So, my question is, **in the problem of relative motions when should we consider that both of the entities are moving or one of them is moving?**

Thank you

Huh, can you share the link for the first one.

Did Greg calculate the time by subtracting 5miles/hour from both of them and making Frank stationary? Or did he just make him stationary without subtracting 5 from Frank as well?

To answer your question, unless the problem explicitly says otherwise, we always consider both entities to be in motion.

In the second question, the bus breaks down, hence we know that it has stopped moving, in the first question there is no such indication

Yes, in the first question after 2 hours Greg considered that Frank had stopped moving but nowhere in the question it is **explicitly** mentioned. Here the link, the question starts at 57:18

No, but he is taking it into account by subtracting his speed from John.

Here is a longer explanation for of why we can do this:

Let the time taken by John to meet Frank be t hours

Then distance b/w starting point and meeting point will be:

(From John’s POV):

d = 10 mph x t hours = 10t miles

(From Frank’s POV):

d = 5 mph x (t+2) hours = (5t + 10) miles

As distance is the same for both of them:

10t = 5t + 10, which is the same as:

10t - 5t = 10

**(10 - 5)t = 10**

5t = 10

t = 2 hours

The relative speed setup allows us to jump to this equation without all of this setup

And no, Greg does not say that the question is making Frank still, he is deducting Frank’s speed from John’s, hence he is taking Frank’s motion into account.

Here is how the question would be solved had Frank stopped.

Distance traveled by John = 10t

Distance traveled by Frank = 5*2 = 10

10 t = 10

t = 1 hour

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