Math Word Problems Demystified - A Self Teaching Guide

It will take 0.5 hour or half an hour to overtake the person on the

bicycle.

Another type of distance problem involves an airplane flyingwith or

againstthe wind or a boat movingwithoragainstthe current. If an airplane is

flying in aheadwind, the speed of the airplane is slowed down by the force of

the wind. If an airplane is flying in atailwind, the speed of the airplane is

increased by the wind. For example, if an airplane is flying at an airspeed of

150 miles per hour and there is a 30 mile per hour tailwind, then the ground

speed of the airplane is actually 150þ 30 ¼180 miles per hour. The airspeed

is the speed of the plane as shown on its speedometer, but if you were

standing on the ground, you would clock the speed at 180 miles per hour. If

the plane had an airspeed of 150 miles per hour and it was flying in a

headwind of 30 miles per hour, the ground speed of the airplane would be

150 
 30 ¼120 miles per hour. In order to solve these problems using algebra,

the direction of the wind must be parallel to the destination of the airplane.

When it is not, trigonometry must be used.

In a similar situation, if a boat is moving downstream at 25 miles per hour

(indicated on its speedometer) and the current is 3 miles per hour, then the

actual speed of the boat is 25þ 3 ¼28 miles per hour since the current is

actually pushing the boat. If the boat is going upstream against the current,

then the current is pushing against the boat and holding it back. In this case,

the speed of the boat is 25
 3 ¼22 miles per hour.

EXAMPLE:A boat’s speedometer reads 20 miles per hour going down-

stream and it reaches its destination in^34 of an hour. If the return trip takes

one hour at the speed of 20 miles per hour, how fast is the current?

SOLUTION:

GOAL: You are being asked to find the speed (rate) of the current.

STRATEGY: Letx¼the rate of the current; then the speed of the boat

downstream is 20þxand upstream is 20
x. The times are given.

Rate  Time ¼ Distance

Downstream 20 þx^3

4

3

4

ð 20 þxÞ

Upstream 20 
x 1 1(20
x)

Since the distances are equal, the equation is^34 (20þx)¼1(20
x).

140 LESSON 13 Solving Distance Problems