I have a problem I haven't been able to solve for a class. A man took a trip in a car. He drove $70$ miles at a slower speed. Then, he went the next $300$ miles at a speed that was $40$ mph faster than earlier. The time he spent driving at the faster speed was twice the time spent at the slower speed. Find the two speeds. I think I could break this down to the following.
Let $s_1$ be the slower speed, $s_2$ the faster speed, $d_1$ the shorter duration, and $d_2$ the longer duration. Then
$d_1=\frac{70}{s_1}$
$d_2=\frac{300}{s_1+40}$
$d_2=2d_1$
and so
$2d_1=\frac{300}{s_1+40}$
I think in my head I can come up with $35$ mph and $75$ mph but that's just because I tried a bunch of numbers that seemed normal for driving. How can I solve this?
Best Answer
Let $t_1$ be the time spent during the first part and $v_1$ be the speed of the first part
We have :
$t_1 \times v_1 = 70 $
$2t_1 \times (v_1 +40) = 300$
So by replacing $t_1 \times v_1$ in the second equation we deduce $ t_1 = 2 $.
And then we find $v_1 = 35$.