# Physics/maths teacher having trouble figuring out this question.

• April 7th 2013, 04:37 PM
vjnr
Physics/maths teacher having trouble figuring out this question.
Hi all

I have a question that even a maths/physics teacher having problem solving, does anyone have any idea how to work is out?

Question:
A line of Army men starts marching along the road. The length of the line is X metre and is moving at a constant speed. A Sergeant, doing inspections, begins at the rear of the line, marches forward toward the front of the line. When he reaches the front of the line, he promptly turns around, and marches back to the rear of the line. While the sergeant was performing his inspection, the line moved X metre. Assuming that all people travel at constant speeds, what was the ratio between the speed of the sergeant and the speed of the Army line?

Thanks
Van
• April 7th 2013, 05:24 PM
topsquark
Re: Physics/maths teacher having trouble figuring out this question.
Quote:

Originally Posted by vjnr
Hi all

I have a question that even a maths/physics teacher having problem solving, does anyone have any idea how to work is out?

Question:
A line of Army men starts marching along the road. The length of the line is X metre and is moving at a constant speed. A Sergeant, doing inspections, begins at the rear of the line, marches forward toward the front of the line. When he reaches the front of the line, he promptly turns around, and marches back to the rear of the line. While the sergeant was performing his inspection, the line moved X metre. Assuming that all people travel at constant speeds, what was the ratio between the speed of the sergeant and the speed of the Army line?

Thanks
Van

We need more information about the time it takes Sarge to travel both of the distances (up or down the line.)

Moving up the line:
$v_s - v = \frac{X}{T_1}$
where v_s is the speed of the Sergeant, v is the speed of the soldiers and T_1 is the time it takes for the Sergeant to get to the front of the line.

Likewise, going back down the line:
$v_s + v = \frac{X}{T_2}$
where T_2 is the time to get to the back of the line.

Solving the top equation for X, subbing this into the bottom equation, then solving for v_s/v gives me
$\frac{v_s}{v} = \frac{T_1 + T_2}{T_1 - T_2}$

So if we know the T's...

-Dan
• April 7th 2013, 07:59 PM
ibdutt
Re: Physics/maths teacher having trouble figuring out this question.
Let the speed of soldiers be u . so the speed of the Sargent = ku
Now when the Sargent goes towards the front his relative sped would be ( ku – u) so the time taken would be X/(ku-u) and when he goes backwards his effective speed would be ( ku+u) and so the time taken would be X/(ku+u) Thus total time taken got going forward and coming backward = X/(ku-u) + X/(ku+u) --- 1
During this time the line of soldiers moves distance = X Time taken will be X/u ----2
Now we can form one equation and simplify to get the value of k that is the ratio of the speeds of Soldiers and the Sargent.