Word Problem - Linear distance in terms of a walking velocity and a second driving velocity

Question

Every day I pick up my son from the bus stop at $4:00$ PM and go home. One day he started walking (towards home), leaving school at $3:30$ and I met him at some point in between. We arrived home $10$ minutes earlier then usual. How far did my son walk? This problem seems to be easily solved with algebra but intuitively it seems the son walks a third as fast as I drive making the $10$ minutes to $30$ minutes ratio $\frac{1}{3}$. That would mean it usually takes $15$ minutes to drive one way. Can anyone describe what's going on?

Answer 1

Examining your word problem, one thing goes through my mind:

So many variables, so little time.

It might be the case that the solution is not unique!

OK, let us try to crank out any solution.

Suppose the distance of the school from home is 30 miles and the car velocity is given by

$\tag 1 v_c = \frac{1 \text{ mile}} {\text{min}}$

and you leave the home at 3:30 PM to pick up your son.

Before your son started leaving early, it took you 30 minutes to get to the school bus stop and 30 minutes to get home, so you got home at 4:30 PM, so a 60 minute round-trip.

Suppose now that your son starts walking to meet you, leaving also at 3:30 PM at a velocity of $v_s$. We are asked to calculate the distance $P$ where you meet your son from the school bus stop ($S = 0$) with the home at $H = 30$:

Now we know that time is distance divided by velocity, and the round-trip is now 50 minutes,

$\tag 2 \frac{H-P}{1} + \frac{P}{v_s} = 50$

and

$\tag 3 \frac{P}{v_s} = \frac{H-P}{1}$

So $2H - 2P = 50$. Since $H = 30$, $P = 5$.

---

There can be many solutions to this problem:

Suppose again the distance of the school is 30 miles but now you have a really fast car that travels at a velocity of $\frac{3 \text{ mile}} {\text{min}}$. In this case you leave the house a 3:50 PM and get home at 4:10 PM when your son doesn't walk. If your son walks a distance of 5 miles, he can cut down your round trip to 50 miles. To travel 50 miles takes your car

$\tag 4 \frac{50}{3} = 16.6666$

If you leave at 3:50 PM you both get home at around 4:07 PM, so your son's effort only saved you $ 3 \frac{1}{3}$ minutes, not 10 minutes. So the distance your son travels is not uniquely defined in this word problem.

If your son can save you 10 minutes, the variables are constrained and related in some way:

What if the bus stop is right at the front of your house? How long does it take to back your car out of the driveway?

Answer 2

You must have picked him up at $3:55$ pm, $5$ minutes away from the bus stop at the speed you travel, thus saving $10$ minutes on the round trip. He was walking for $25$ minutes so your speed is $5$ times faster than his.

Example: Say on a normal day you leave home at $3:45$ pm, drive for $15$ minutes pick your son up at $4:00$ pm, drive for $15$ minutes back and arrive home at $4:15$pm

Next day, you leave home at $3:45$ pm, drive for only $10$ minutes where you meet you son walking and pick him up at $3:55$ pm, drive for $10$ minutes back and arrive home at $4:05$ pm, $10$ minutes earlier than usual.

Your son therefore was walking from $3:30$ pm to $3:55$ pm, for a time of $25$ minutes and walked a distance which you drive in only $5$ minutes (you only drove for $10$ minutes instead of $15$). Hence you travel $5$ times faster than your son walks.

We do not have enough information to determine anything else like the actual speeds or distances.

Say the distance $d_{hs}$from home to school is $5$ miles. A $15$ minute drive for you is done at $$v(mph) = \frac{d (\text{miles})}{t(\text{hrs})} = \frac{5}{(\frac{15}{60})} = 20\text{mph}$$

We worked out that your son walks at one fifth your speed which is $4\text{mph}$ so the distance he walks is determined from $d = vt$

$$d_{son} = 4\cdot \frac{25}{60} = \frac{100}{60} = \frac{5}{3}\text{miles}$$ We are dividing the time in minutes by $60$ to get the time in hours as our speed is in miles per hour.