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If a is travelling at 120 mph and is one lap (3/4 mile) behind, and B is travelling 110 mph, how many laps will it take A to catch up?

math word problem
 

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Ralph L. | Algebra I, II, Visual Basic, Beginning C++ tutorAlgebra I, II, Visual Basic, Beginning C...
4.0 4.0 (1 lesson ratings) (1)
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120t - .75 = 110t
 
120t - 110t = .75
 
10t = .75
 
divide both sides by 10:
 
t = .075 hours
 
.075 * 120 =  9 miles
 
since 1 lap is 3/4 mile, so, 9 miles divided by .75 mile per lap is 12 laps.
 
 

Comments

Ralph,
 
The faster car is travelling at 160 laps/hr, so in t=3/40 hr, it will have travelled 12 laps, not 9.  I think you just mislabelled the units---you meant to say 9 miles, maybe?
 
Regards,
Hassan H.
My mistake. Thank you. Yes, I meant 9 miles.
Brad M. | Professional Tutor: Ultra Streamlined Math - Physics - AccountingProfessional Tutor: Ultra Streamlined M...
4.9 4.9 (218 lesson ratings) (218)
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Hey Rosemary -- here's the "play-by-play": PRETEND "A" is 1 mile behind ... at 10 mph closing speed, one mile needs 1/10 of an hour to close ... 120 mph covers 240 "half-miles"/hr OR 480 "qtr-miles"/hr ... three (3) of these "qtr-miles" would be covered 160 times/hr ... 1/10 of an hr travels 16 laps when 1 mile behind ==> the 3/4 mile lag would take 12 laps to catch up ... Best wishes, ma'am :)