Algebra Word Problem #1

Solving a word algebra problem using three different approaches

The following problem is from the second chapter of this book. I’m going to solve it three different ways.

A highway patrolman spots a speeding car. He clocks it at 70 mph and takes off after it 0.5 mile behind. If the patrolman travels at an average rate of 90 mph, how long will it be before he overtakes the car?

Approach 1 - Same Travel Time, Different Distances

Suppose we want to approach the problem in the following way. We look at the situation from the point in time at which the speeding car is 0.5 miles ahead of the cop car and the cop is beginning his pursuit.

First, let’s start with the most straightforward information that we are given. We are told that the speed of the highway patrolman’s vehicle is 90mph and the speed of the speeding car is 70mph.

$s_{c}$ = 90mph

$s_{s}$ = 70mph

We want to figure out the time the cop car traveled until catching up with the speeding car, which we will call $t_{c}$.

$t_{c}$ = travel time of cop car.

Measured from the point in time at which the cop begins his pursuit, the following times are the same:

1) the time traveled by the cop car until the pursuit is over, and

2) the time traveled by the speeding car until the pursuit is over.

So if we defined $t_{s}$ as the travel time of the speeding car, then:

$t_{s} = t_{c}$

The distance traveled by the vehicles from the point in time we are looking at (when the speeding vehicle is 0.5 miles ahead) are not equal. Let’s say that the cop car travels distance $d$.

$d =$ distance traveled by cop car.

The distance traveled by the speeding car would thus be 0.5 miles less than $d$.

$d - 0.5 = $ distance traveled by speeding car from point in time at which speeding car is 0.5 miles ahead of cop car until it is intercepted by cop car.

Distance divided by rate equals time. We know that the times are equal. We’ve figured out the distances traveled, and we were given the rates of travel (the speeds). So we set $\frac{d}{r}$ for the cop car and the speeding car equal to each other. Doing this means we are treating the times traveled as equal, which we already know we can assume:

$\frac{d}{90} = \frac{d - 0.5}{70}$

Cross multiply

$70d = 90d - 45$

Further simplify the expression

$45 = 20d$

$d = \frac{9}{4}$

That’s the distance traveled by the cop car, in miles. We know as general background knowledge that distance divided by rate equals time. So let’s divide the distance traveled by the cop car by the rate the cop car traveled in order to get $t_{c}$, which was our original goal.

$\frac{\frac{9}{4}miles}{90\frac{miles}{hour}} \rightarrow \frac{9}{4}miles \cdot \frac{1}{90} \frac{hours}{mile} \rightarrow \frac{1}{40}hours $

So $t_{c} = \frac{1}{40}hours$

Approach 2 - Different Travel Times, Same Distances

Suppose you want to approach the problem as follows: you want to initially determine the time the speeding car travels when measured from the time it and the cop car are at the same point on the road, and then work your way to what the book asks (the time the cop car travels when the speeding car is 0.5 miles ahead) from there.

In this setup, the distances traveled by the vehicles would be the same, but the times traveled would be different (because the cop car doesn’t travel, initially, for some period of time up until the speeding car is 0.5 miles ahead).

So let’s define the distance traveled by either car as $d$.

$d = $ distance traveled by either car

Again, we already know the rates they traveled out. So the tricky thing here is the time traveled.

In this setup of the problem, the speeding car’s travel time is measured from when the speeding car and the cop car are at the same point on the road. Let’s call the speeding car’s travel time $t_{s2}$

$t_{s2}$ = speeding car’s travel time when measured from the point at which it and the cop car are at the same point on the road. (Note this is a different travel time than $t_{s}$ in the previous approach!)

Now, the cop car is initially idle while the speeding car is traveling, and the speeding car thus gets a head start. How long is the cop car idle for?

We can figure this out by using the information we are given. We are told that the speeding car’s speeding is 70 miles per hour and that it has traveled 0.5 miles before the cop car begins pursuit. In other words, we have the rate (speed), and we have the distance. Distance divided by rate equals time. If we divide the distance traveled by the speeding car before the cop’s pursuit begins by the rate at which the speeding car was traveling, we can get the time the speeding car spent traveling before being pursued by police.

$\frac{0.5 miles}{70mph} = \frac{0.5}{70}hours$

If we multiply by $\frac{2}{2}$ to get rid of the decimal, we get:

$\frac{1}{140}hours$

Again, this represents the time that the speeding car traveled from the time it was at the same point as the cop car on the road until the point at which the cop car began pursuit. The reason we want this is that we defined $t_{s2}$ as the speeding car’s travel time, and we want to define the cop car’s travel time in terms of that.

$t_{s2}$ = speeding car’s travel time.

$t_{c}$ = cop car’s travel time.

$t_{c}$ in terms of $t_{c}$ = $t_{s2} - \frac{1}{140}hours$

So now we have times traveled by both vehicles expressed in $t_{s2}$ and we were given the rates at which the vehicles traveled. We know the distances are equal. As a general proposition, distance = rate * time. So we can set our rate * times for our two cars equal to each other.

$t_{s2} \cdot 70mph = (t_{s2} - \frac{1}{140}) \cdot 90mph$

$70t_{s2} = 90t_{s2} - \frac{90}{140}$

$ \frac{9}{14} = 20t_{s2}$

$ \frac{9}{280} = t_{s2}$

Ok, so we have a value for the time the speeding car traveled, but the book asked for the time the cop car traveled, so how do we get that from what we have? Well, we defined $t_{c}$ in terms of $t_{s2}$ as follows:

$t_{c} = t_{s2} - \frac{1}{140}hours$

So we just need to plug in our value for $t_{s2}$ into that, and we get $t_{c} = \frac{7}{280}$ which equals $\frac{1}{40}$ hours, which was the same as our first answer.

Approach 3 - Dividing the difference in distance by the difference in speed

At the time the cop begins pursuit, the speeding car has a head start of 0.5 miles. Let’s call this $h$.

$h = 0.5 miles$.

The cop car is gaining on the speeding car at a rate of 20mph (90mph - 70mph = 20mph).

$s_{g} = 20mph$

So we have a distance, $h$, and we have a rate, $s_{g}$.

$\frac{d}{r} = t$. So let’s divide the distance by the rate and figure out how long it takes the cop car to catch up.

$\frac{0.5miles}{20mph} = \frac{0.5}{20}$hours

And let’s multiply the top and bottom by 2 to get rid of the decimal

$\frac{1}{40}$hours