Catch-up problems in Rate, Time, Distance (RTD) category.

Question

Car-X is $40$-miles West of Car-Y. Both cars are traveling East, and Car-$X$ is going $50$% faster than Car-$Y$. If both cars travel at a constant rate, and it takes Car-$X$ $2$ hours and $40$ minutes to catch up to Car-$Y$, how fast was the Car-$Y$ going?

This problem can be solved by using the so-called $(A-B)$ technique as follows:

$1.5r-r = 0.5r$

So, $0.5r \times \frac{8}{3} = 40$
$\implies r = 30$

.

Can the following problem be solved using the above technique?

Scott starts jogging from point-$X$ to point-$Y$. A half-hour later his friend Garrett who jogs $1$ mile per hour slower than twice Scott's rate, starts from the same point and follows the same path. If Garrett overtakes Scott in $2$ hours, how many miles will Garrett have covered?

If No, why not?

Answer 1

You can use the same approach

Let us take Garrett's rate $=(2x-1)$ mph and

Scott's rate $=x$ mph

Since both are walking in same direction, speed$=2x-1=x-1$

The formula for relative distance between Garrett and Scott $=$ distance covered by Scott in initial half an hour which is $=\frac x2$

Now, $$\frac x2=(x-1)2$$ $$x=4x-4$$ $$x=\frac43mph$$

The number of miles covered by Garrett in $2$ hours is $=(2(\frac43)-1)2=\frac{10}{3}$miles

Answer 2

Let Garrett's speed be $v$ miles per hour.

Scott's speed is half of a speed that is $1$ mph faster than Garrett's, that is, Scott's speed is $\frac12(v + 1).$

The rate at which Garrett closes the distance to Scott (after Garrett starts running) is $ v - \frac12(v + 1) = \frac12(v - 1).$

In $\frac12$ hour, Scott covers the same distance at rate $\frac12(v + 1)$ that Garrett is able to close in $2$ hours at rate $\frac12(v - 1)$. That is, $$ \frac12\left(\frac12(v + 1)\right) = 2\left(\frac12(v - 1)\right). $$ Simplifying, $$ \frac14(v + 1) = v - 1, $$ $$ \frac54 = \frac34v, $$ $$ v = \frac53.$$

So in $2$ hours, the distance Garrett covers is $$ 2 \left(\frac53\right) = \frac{10}{3}. $$

This is not a better solution than the other way; I just wanted to demonstrate that it is not necessary to solve for Scott's speed explicitly. With a little more effort we could avoid explicitly solving for Garrett's speed as well!