Two cars are traveling on a highway in the same direction. If car A

This is a "runaway" problem with a "gap" and a trap (gap distance that Car A must travel is not 36 miles).

Use RT=D and "solve for the gap."
We need
1) the rate at which the gap widens ("relative speed")
2) gap distance

(1) Rate = relative speed. Travel is in SAME direction. Subtract slower speed from faster. Relative speed: \((Rate_{A}-Rate_{B})=(55-45)=10mph\)

(2) Gap distance?
The gap distance at start: 18 miles
Desired gap distance? (18*2) = 36 miles
Car A is faster than B and ahead of car B.
The 18-mile distance between them never shrinks. Rather, that gap increases slowly.

Car A needs to get (36-18) = 18 ADDITIONAL miles ahead of Car B
\(D_{gap}=18\) miles

(3) TIME required?
How long will it take for Car A to put another 18 miles between A and B?

Relative speed, \(R=10\) mph
Distance of gap, \(D=18\) miles
Time needed: \(RT=D\), so \(T=\frac{D}{R}\)
\(T=\frac{18}{10}=\frac{9}{5}=1\frac{4}{5}\) hours

Multiply any fraction of an hour by 60 to get minutes: \((\frac{4}{5}*60)=48\) minutes

Time needed: 1 hour, 48 minutes

ANSWER A



Check. From the time that A was 18 miles ahead of B to the time that A was 36 miles ahead of B, both cars traveled for \(\frac{9}{5}\) hours.

Start: Car B was at the 0 mile mark
RT=D, so Car B traveled \(45*\frac{9}{5}=81\) miles
End: Car B is at the 81-mile mark

Start: Car A started at the 18-mile mark
Car A traveled \(55*\frac{9}{5}=99\) miles
End: Car A is at mile marker (18+99) = 117

Distance between A and B? (117-81)=36 mi

That works.