jennysussna wrote:
Two cars are traveling on a highway in the same direction. if car A traveling at a rate of 55mph is 18 miles ahead of car B, which is traveling at 45 mph, how muh time will it take for car a to double the distance between itself and car b?
a 1 hour and 48 mins
b 3 hours
c 3 hours and 36 mins
d 4 hours
e 4 hours and 18 mins
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) =
117Distance between A and B? (117-81)=36 mi
That works.
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