SOLUTION: Hi! I've become stuck on a section in this word problem that I can't seem to figure out. Can you assist me? A woman driving a car 14 ft long is passing a truck 28 ft long. Th

Algebra ->  Equations -> SOLUTION: Hi! I've become stuck on a section in this word problem that I can't seem to figure out. Can you assist me? A woman driving a car 14 ft long is passing a truck 28 ft long. Th      Log On


   



Question 196265: Hi!
I've become stuck on a section in this word problem that I can't seem to figure out. Can you assist me?
A woman driving a car 14 ft long is passing a truck 28 ft long. The truck is traveling at 50 mi/h. How fast must the woman drive her car so that she can pass the truck completely in 6 s, from the position shown in figure (a) to the position shown in figure (b)? [Hint: Use feet and seconds instead of miles and hours.] (Enter your answers in terms of integers or fractions.)
thanks!

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!

A woman driving a car 14 ft long is passing a truck 28 ft long. The truck is traveling at 50 mi/h. How fast must the woman drive her car so that she can pass the truck completely in 6 s, from the position shown in figure (a) to the position shown in figure (b)?
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50 mph(1 hr/3600 sec)(5280 ft/1 miles) = 73 1/3 ft/sec
In 6 seconds the front/back of the bus goes 440 ft
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The back of the car must clear the front of the bus.
The back of the car must go 14+28+440 ft = 482 ft in 6 seconds
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Rate of the Car:
r = d/t
rate = (482 ft)/(6 sec) = 80 1/3 ft/sec
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(241/3 ft/sec)(3600 sec/1 hr)(1 mile/5280 ft) = 54.77 mph.
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Cheers,
Stan H.