Word problem: Finding The Length Of A Column Of Marching Soldiers

[NotaMathPerson]

An army of soldiers is marching down a road at 5 mi/hr. A messenger on horseback rides from the front to the rear and returns immediately, the total time taken being 10 minutes. Assuming that the messenger rides at the rate of 10mi/hr, determine the distance from the front to the rear.

Can you help me get started with the problem?

Thanks!

 

[MarkFL]

Re: Word problem

We can greatly simplify this problem if we think in terms of the speed of the messenger relative to the marching soldiers. As the messenger rides to the rear of the column, his speed relative to the column is the sum of their actual speeds and when he returns to the front, his relative speed is the difference of their actual speeds. So, we have 3 unknowns: the length of the column (the distance $d$ from the front to the rear), the time $t_1$ spent traveling to the rear and the time $t_2$ spent returning to the front.

Can you now set up 3 equations including these 3 unknowns? Two of the equations will involve the relationship between distance, velocity and time, while the third relates to the relationship between $t_1$, $t_2$ and the total time traveled (which you will want to express in hours since the speeds are given in miles per hour).

 

[NotaMathPerson]

Re: Word problem

We can greatly simplify this problem if we think in terms of the speed of the messenger relative to the marching soldiers. As the messenger rides to the rear of the column, his speed relative to the column is the sum of their actual speeds and when he returns to the front, his relative speed is the difference of their actual speeds. So, we have 3 unknowns: the length of the column (the distance $d$ from the front to the rear), the time $t_1$ spent traveling to the rear and the time $t_2$ spent returning to the front.

Can you now set up 3 equations including these 3 unknowns? Two of the equations will involve the relationship between distance, velocity and time, while the third relates to the relationship between $t_1$, $t_2$ and the total time traveled (which you will want to express in hours since the speeds are given in miles per hour).

Hello! I am not sure about this but this is how I will attack the problem based on the relative speed.

Let
x = time taken to travel front-rear
1/6-x = time taken to travel rear-front

Since the distance traveled for two trips are equal

$15x = 5(\frac{1}{6}-x)$

$x = \frac{1}{24}$ hr.

$D = \frac{5}{8}$ mi

Is this correct. If not, please tell me why. Thanks!

 

[MarkFL]

Let's work this problem in general terms so that we have a formula we can plug the given data into, so that if presented with s similar problem, we won't have to work essentially the same problem all over again.

So, let's let:

$$v_S$$ = the speed of the soldiers.

$$v_M$$ = the speed of the messenger, where $v_M>v_S$.

Now, we may state:

$$d=\left(v_M+v_S\right)t_1$$

$$d=\left(v_M-v_S\right)t_2$$

$$t_1+t_2=t$$

And so we obtain:

$$\left(v_M+v_S\right)t_1=\left(v_M-v_S\right)\left(t-t_1\right)=\left(v_M-v_S\right)t-\left(v_M-v_S\right)t_1$$

$$\left(v_M+v_S\right)t_1+\left(v_M-v_S\right)t_1=\left(v_M-v_S\right)t$$

$$2v_Mt_1=\left(v_M-v_S\right)t$$

$$t_1=\frac{\left(v_M-v_S\right)t}{2v_M}$$

Hence:

$$d=\left(v_M+v_S\right)\cdot\frac{\left(v_M-v_S\right)t}{2v_M}=\frac{v_M^2-v_S^2}{2v_M}t$$

Plugging in the given data, we find:

$$d=\frac{\left(10^2-5^2\right)\left(\dfrac{\text{mi}}{\text{hr}}\right)^2}{2\cdot10\left(\dfrac{\text{mi}}{\text{hr}}\right)}\cdot\left(10\text{ min}\cdot\frac{1\text{ hr}}{60\text{ min}}\right)=\frac{75}{120}\text{ mi}=\frac{5}{8}\text{ mi}\checkmark$$