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In the following and similar uniform motion problems, when equating unequal times, the additional time seems to be added to the faster object in order to solve the problem correctly.

However, to me this seems unintuitive as a $r = \frac Dt$ implies a faster object should take less time. Nevertheless, adding the time in this way seems to be the only way to solve the problem.

I do not understand why this is the case.

The problem is as follows:

A man cycles downhill 12 miles from his home to the beach and then later cycles back to his home. The man's speed returning home uphill is 8 mph slower than his downhill speed. He also takes 2 hours longer to return home than it took him to get to the beach. What is the man's speed cycling downhill?

(Question sourced from OpenStax; Intermediate Algebra, pp. 715-716, https://openstax.org/details/books/intermediate-algebra)

I believe I understand most of the principles to solve this problem.

We know the distance, we know the distance is the same in both directions, but we don't know the rate or time of travel in either direction.

We choose to equate the time of travel because this will provide an equation in one variable; equating distance would result in an equation in two variables.

We therefore equate the two expressions of time based on $t = \frac Dr$.

Let $r$ be the rate of travel downhill.

The expression of time for the man travelling downhill would be $\frac{12}{r}$ and the expression of time for travel uphill would be $\frac{12}{r-8}$.

To equate the two expressions, we need to include the fact that the travel uphill was two hours longer than the travel downhill, i.e. the uphill travel is equal to the downhill travel plus two.

Hence, we have for the expression of time for the uphill travel: $\frac{12}{r-8} + 2$.

The equation I end up with is therefore: $$\frac{12}{r} = \frac{12}{r-8} + 2$$.

However, solving this leads me to a quadratic equation that I don't think can be factored. $$2(r^2-8r+48)$$.

The correct equation is: $$\frac{12}{r} + 2 = \frac{12}{r-8}$$

But here the extra 2 hours are added to the expression for the downhill travel.

I do not understand why the extra 2 hours are being added to the expression of time for downhill travel. The downhill travel is faster and to add 2 hours of time would imply it is 2 hours slower.

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The mistake is in the expression for time of uphill travel.

Suppose we have $T_u = {12 \over r - 8}$ as uphill travel time, and $T_d = {12 \over r}$ for downhill. From the problem we know that traveling uphill takes 2 hours longer than downhill i.e. $$ T_u = T_d + 2 $$ So the equation should instead be $$ {12 \over r - 8} = {12 \over r} + 2 $$

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Suppose $T$ be the bigger time interval and $t$ be the smaller time interval.We can write $T-t=\Delta t\Rightarrow T=t+\Delta t$. So as you see, the time difference got added to the smaller time interval.
In other words, you can't equate a smaller time interval to a bigger time interval. To do this you have to add something to the smaller time interval to make it big enough to be equated to the bigger time interval.

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