The Ultimate Bus

Buses are a nation's pride in some countries and a disgrace in others. Math questions can handle random behavior like the bus schedule in some countries, but it's much better to deal with a good bus system, with a normal schedule. Here is one question about such system.

Andy walks at a constant speed along a long street. Every 6 minutes he is passed by a bus going in this direction. Every 2 minutes he is also passed by a bus going in the opposite direction. Both directions have the same schedule and all buses go at the same speed. What is the time difference between two consecutive bus departures?

First we need to define some variables and understand a conecpt of of physics. Let's call the speeds of Andy and the buses a and b, respectively.

A relative speed is the speed at which two distinct objects move towards or away from each other. This speed is given by subtracting the two speeds. In the first case, the bus is approaching Andy at a speed of b - a. Since the other buses go in the opposite direction, their speed becomes "negative", and the bus goes towards Andy at a speed of b + a.

Let x be the distance between the buses (they are scheduled on regular intervals, so the distance between buses is the same for all buses)

Using the formula distance = velocity * time, we can say that:
x = (b - a) * 6
x = (b + a) * 2

Equate the x's:
6b - 6a = 2b + 2a
4b = 8a
b = 2a
a = b/2

Either Andy is very fast of the buses are very slow, because the buses go at a speed just twice as Andy. That is definitely weird.

x = (b - b/2) * 6 = b / 2 * 6 = 3b

Since the buses need to travel a distance of 3b to get from the location of one bus to another bus and they do so at a speed of b, a bus must be sent every 3b / b = 3 minutes. That's a great interval for the passengers of this bus system.

Hope you liked it,
Nadav

nadavs

Towns With Angles

Today's question also involves the law of cosines, but with a different purpose. It is a very neat question from a test which is considered "hard". Try it before looking at the answer.

A biker travels from town A to town B at 10 km/h. Another biker is traveling from town B to town C at 12 km/h. The distance from town A to town B is d. Angle ∠ABC = 120°. The bikers are closest after 2.5 hours of riding. Find d.

First, try to draw the situation. Draw the three towns, draw d, draw the angle, and place two points on segment AB and segment BC, which are the bikers.

Let t be the time the bikers are traveling. This means the first biker's distance from town A is 10t and the second biker's distance from town B is 12t. Since the first biker's distance from town A is 10t, his distance from town B is d - 10t.

We need a variable that represents the distance between the bikers. Let's call it x.

As you can see now, we have a triangle with three sides (x, 12t, d - 10t) and an angle of 120° which is opposite to the side with measure x. This situation calls for the law of cosines to find the relation of all variables:
x² = (12t)² + (d - 10t)² - 2(12t)(d - 10t)cos 120°

cos 120° = -0.5, so the relation becomes (after expanding):
x² = 144t² + d² - 20dt + 100t² + 12dt - 120t²

Add like terms and take the square root:
x = √(124t² + d² - 8dt)

We know that x has a minimum when t = 2.5, so if we derive x and plug 2.5 for t, we can find d.

As you should know, when you derive √f(x), you get f'(x)/2√f(x). Since we equate the derivative to zero (at a maximum or a minimum, the derivative is zero), we can now ignore the denominator (since it's always positive and we can multiply by it). Don't ignore it on a test, I'm doing it to save time.

Since we derive by t (dx/dt), the derivative of d² is 0 (d is a constant, not a variable).
0 = 248t - 8d

We know that t = 2.5, so:
0 = 620 - 8d
8d = 620
d = 77.5 km

Towns A and B are 77.5 km apart.

Have a great weekend,
Nadav

nadavs

Boating Calculus

Today's question is dedicated to all boating lovers and calculus fans. It is not very hard, it has no use in real life, and yet it is a very beautiful question (like the ones on super math tips).

A boat leaves the dock at 12:00pm and heads west at 40km/h. Another boat travels north at 20km/h and enters the dock at 1:00pm. When is the distance between the boats the shortest and what is that distance?

Nice, isn't it? We have an optimization question here, so we need to create a function to find a variable to optimize. We're dealing here with given speeds, variable times and varying distances, so let's call the time variable t (in hours, to match the speeds) and the distance variable d (in km).

Now, the distance from the first boat to the dock is 40t (it starts at the dock). The distance from the second boat to the dock is 20 when it starts, and 20 - 20t total (since it travels towards the dock).

The boats are heading in perpendicular directions, so they form a right triangle. Using the Pythagorean theorem, we can find the distance between the boats, d:
d² = (40t)² + (20 - 20t)²
d² = 1600t² + 400 - 800t + 400t²
d² = 2000t² - 800t + 400

Since we want to find the minimum value for d and d must be positive, we can also find the minimum value of d² (also must be positive). Let y = d²:
y = 2000t² - 800t + 400

Now differentiate this function:
y' = 4000t - 800

To find an extremety, we need to set y' to zero:
0 = 4000t - 800
4000t = 800
t = 1/5

To see that it is indeed a minimum, let's find the second derivative of y:
y'' = 4000
Since it is positive, y is minimal when t = 1/5

Now calculate y when t = 1/5:
y = 2000 * (1/5)² - 800 * (1/5) + 400
y = 2000 * (1/25) - 160 + 400
y = 80 + 240 = 320

However, y is the distance squared, so let's take the square root of this number:
d = sqrt(y) = sqrt(320) = 8sqrt(5) = 17.888... km.

This means that after 1/5 of an hour, which is 60 / 5 = 12 minutes, the boats will be the closest to each other at 8sqrt(5) (or 17.888) km apart.

Hope you enjoyed.
Nadav

nadavs