Trajectory/velocity/time question

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The discussion revolves around a physics problem involving a tourist walking alongside a moving vehicle carrying a log. The tourist's velocity is 2 m/s, and he counts 16 steps (12 m) from the front to the back of the log and 112 steps (84 m) from the back to the front. Using the equations for relative velocity, the calculated speed of the vehicle is 1.5 m/s, and the length of the log is determined to be 21 m. The participant expresses confusion about the derivation of the second equation but receives clarification on how to approach the problem. The collaborative effort leads to a successful resolution of the homework question.
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Homework Statement


(I'm translating the problem from Slovak, so it might sound a bit choppy at times-my apologies)
There's a road. On the road, there's a moving vehicle carrying a tree log, and a tourist is walking alongside the vehicle. The tourist's velocity is 2 m/s, and he wants to know how long the log is. When he walks from the front end to the back end of the log (keep in mind the log is moving in the opposite direction) he counts 16 steps. When walking from the back end to the front end of the log, he counts 112 steps (the vehicle is now moving in the opposite direction than the tourist). The length of his step is 0.75 m.
a) What is the velocity of the vehicle?
b) What is the length of the log?


Homework Equations


s=t.v
s...trajectory
t...time
v...velocity


The Attempt at a Solution


I'm rather stumped. I converted the tourists steps:
It took him 12m and 6s to walk from the front to the back end of the vehicle.
It took him 84m and 42s to walk from the back to the front end of the vehicle.
And that's about as far as I got. I could calculate the vehicle velocity if I had the log length, just like I could calculate the log length if I had the velocity. But having neither?

I also deduced he must be moving quicker than the vehicle, because he managed to reach the front end from the back one.

Thanks for any help.
 
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Welcome to PF.

It seems like the length of the log can be expressed as

T1*(Vp + Vt) = L

And T2*(Vp - Vt) = L

Where Vp is the velocity of the person, Vt the Velocity of the truck.

T1 and T2 you already figured looks like.
 
So, I put together the equation by combining yours, and calculated the thing. My results are
a) The speed of the vehicle is 1,5 m/s.
b) The log is 21 m long.

Can anyone check my results? Thanks.

Also, though I get how you got the first equation (since they're moving in the same direction, it's just time multiplied by their combined speed), I don't understand how I'd put together the second one. Can you explain? Thankies, and thanks so much for your help-I thought I was never going to solve it!
 
Puchiko said:
So, I put together the equation by combining yours, and calculated the thing. My results are
a) The speed of the vehicle is 1,5 m/s.
b) The log is 21 m long.

Can anyone check my results? Thanks.

Also, though I get how you got the first equation (since they're moving in the same direction, it's just time multiplied by their combined speed), I don't understand how I'd put together the second one. Can you explain? Thankies, and thanks so much for your help-I thought I was never going to solve it!

That's what I get anyway.

In the one you are walking against the speed of the log. So in relative terms the end of the log is approaching the walker at the combined rates of their speeds.

In the other he is walking toward the front of the truck and the front end of the log at the rate of his speed less the speed that the truck is moving since it's moving in the same direction he is.

In both he walks the length of the log at the effective rate in the time given by his walking speed and length of stride.
 
Thanks for all your help, it is appreciated.
 
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