SOLUTION: A woman driving a car 14 ft long is passing a truck 30 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 i
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Question 158128This question is from textbook Algebra and trigonometry
: A woman driving a car 14 ft long is passing a truck 30 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)?
(A)
---CARI --->
---------I
---------IT-R-U-C-K -->
---------I
(B)
-------------I
-------------ICAR -->
-------------I
T-R-U-C-KI ---> This question is from textbook Algebra and trigonometry
You can put this solution on YOUR website! A woman driving a car 14 ft long is passing a truck 30 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)?
She would really have to be flying to pass that truck in
only 6 seconds.
First we must change 50 mi/h to ft/s.
replace by
replace by
replace by
So in 6 seconds, using ,
the truck travels
We start with the front bumper of the car even
with the tailgate of the truck.
So in that 6 seconds, the front bumper of the
car must travel:
1. the 440 feet which the truck travels in 6 seconds.
2. the 30 feet from the tailgate of the
truck to the front bumper of the truck.
3. the 14 feet to get the back bumper of
the car even with the front bumper of the
truck.
That's a total of feet, which she
must travel in 6 second.
Now we find her rate using
Now we must get that to :
Multiply by ×
Multiply by ××
Multiply by ×××
Cancel the , the , and the :
×××
That's fast!!! I don't she'll do it, do you?
Edwin
You can put this solution on YOUR website! 264,000 per hour
264000ft/3600 sec
(264,000/3600)=(x/6)
x=440 feet+16+30=484 ft/6sec
290400ft/3600 sec
55 miles per hour