Question

Loren drove 200 miles at a certain rate, and his wife, Lois, drove 100 miles at a rate 10 mph slower. If Loren had driven for the entire trip, they would have arrived 30 minutes sooner. What was Loren's rate?

Answer 1

As long as Loren drove, the law of motion was

$200 = st_1 \implies t_1 = \dfrac{200}{s}$

As long as Loid drove, the law of motion was

$100 = (s-10)t_2 \implies t_2 = \dfrac{100}{s-10}$

So, the total time they took is

$t_1+t_2=\dfrac{200}{s}+\dfrac{100}{s-10}$

Had Loren driven the whole time, the law of motion would have been

$300=st_3 \implies t_3 = \dfrac{300}{s}$

And we know that this time would have been 30 minutes (i.e. 0.5 hours) faster. So, we have

$t_3 = t_1+t_2-0.5$

This translates into

$\dfrac{300}{s}=\dfrac{200}{s}+\dfrac{100}{s-10}-\dfrac{1}{2}$

If we subtract 200/s from both sides, we have

$\dfrac{100}{s}=\dfrac{100}{s-10}-\dfrac{1}{2}$

We can simplify the right hand side by summing the two fractions:

$\dfrac{100}{s-10}-\dfrac{1}{2} = \dfrac{200-(s-10)}{2(s-10)}=\dfrac{210-s}{2(s-10)}$

So, we have to solve

$\dfrac{100}{s}=\dfrac{210-s}{2(s-10)}$

If we cross multiply the denominators, we have

$200(s-10)=s(210-s) \iff 200s-2000=210s-s^2 \iff s^2-10s-2000=0$

Which yields the solutions

$s=-40,\quad s=50$

We accept the positive solution, because the negative would mean to travel backwards, so Loren's rate was 50mph

Answer 2

Your answer is 50 mph