Question 480042: if I want to pass a car just in front of me going at 55mph, what will my speed be to pass within 1/4 of a mile? Show me the mathematical work thanks.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the car in front of you is going 55 miles per hour.
you are right behind it.
let's say you are 50 feet behind it.
let's say that, after you pass the car, you will be 50 feet in front of it.
let's say that the car is 10 feet in length.
let's say that your car is 10 feet in length.
if your car is 50 feet behind the car in front of you, then the front of your car is 50 feet behind the back of it.
in order for you to get 50 feet in front of that car, you will have to travel 120 feet further than the other car is traveling.
this includes 50 feet from behind that car plus 10 feet of the car plus 50 feet in front of the car plus 10 more feet because you need to get the back of your car to be 50 feet in front of that car.
you want to be able to do this within 1/4 of a mile.
1/4 of a mile is equal to 5280 / 4 = 1320 feet
you need to complete your maneuver within 1320 feet.
this means that at the 1320 foot mark from when you start, you need to have the front of your car touching the 1320 foot mark.
this also means that at the 0 foot mark (when you start your maneuver), you need to have the front of your car touching the 0 foot mark.
you have to travel 1380 feet in the same time that he travels 1260 feet.
we will need to translate miles per hour into feet per second so that we can solve this problem using feet and seconds rather than miles and hours.
he is traveling at 55 miles per hour.
multiply that by 5280 to get feet per hour.
divide that by 60 to get feet per minutes.
divide that by 60 again to get feet per second.
you get 55*5280/60/60 = 80.66666667 feet per second.
for him to travel 1260 feet at 80.66666667 feet per second, it will take him 1260/80.66666667 = 15.61983471 seconds.
you need to travel 1380 feet in 15.61983471 seconds in order to pass him within 1320 feet.
rate * time distance.
your rate is x (unknown at this time).
your time is 15.61983471 seconds.
your distance is 1380 feet.
formula becomes:
rate * 15.61983471 seconds = 1380 feet.
divide both sides of that equation by 15.61983471 to get:
rate = 1380 / 15.61983471 = 88.34920635 feet per second.
to translate 88.34920635 feet per second into miles per hour, you need to divide by 5280 and multiply by 60 and multiply by 60 again to get:
88.34920635 * 60 * 60 / 5280 = 60.23809524 miles per hour.
to confirm, we need to go back to the feet per second calculations.
his rate is 80.66666667 feet per second.
your rate is 88.34920635 feet per second.
in 15.61983471 seconds, he has traveled 80.66666667 * 15.61983471 = 1260 feet.
in the same 15.61983471 seconds, you have traveled 88.34920635 * 15.61983471 = 1380 feet.
you have traveled 120 more feet than he has traveled which is the distance that you needed in order to go from 50 feet behind him to 50 feet in front of him.
this allows the front of your car to be 50 feet behind the back of his car when you start and allows the back of your car to be 50 feet in front of the front of his car when you're done.
check out the picture to see what i mean.
in the car, the box is the car.
the right side of each car is the front of that car.
the left side of each car is the back of that car.
your answer is:
you have to be traveling 60.23809524 miles per hour in order to complete the maneuver within 1/4 of a mile.
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Keep in mind that we are talking about average speed throughout the distance.
to maintain an average speed of 60 miles per hour, you would have to attain speeds in excess of that.
my own calculations indicate that a top speed of approximately 67 miles per hour for a period of approximately 1.57 seconds would allow the average speed throughout the entire distance to be equal to approximately 60 miles per hour.
the devil is in the details and the numbers are not exact, but this should give you an idea.
the assumption was a gradual acceleration to top speed from 55 miles per hour and then a gradual deceleration from top speed back to 55 miles per hour at the finish.
i divided the distance into 9 segments.
each segment was 1380 / 9 = 153.33333333 feet long.
the first and ninth segment was accomplished at an average speed of 55 miles per hour.
the second and eighth segment was accomplished at an average speed of 57.9 miles an hour.
the third and seventh segment was accomplished at an average speed of 61.4 miles per hour.
the fourth and sixth segment was accomplished at an average speed of 64.8 miles per hour.
the fifth segment was accomplished at an average speed of 66.4 miles per hour.
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i specified the speed for all segments except the 5th segment.
the speed in that segment was calculated from the remaining time left after adding up the total time used in the other segments.
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since this was an average speed per segment, the same logic could be used to divide the last segment into smaller segments and would probably yield a top speed in excess of 66.4 miles per hour, but not a lot more.
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it's not exact science, but it should give you an idea.
here's the table of values that i used.
miles per hour
segment number
feet per second
number of seconds
55 1 80.66666667 1.900826446
57.95454545 2 85 1.803921569
61.36363636 3 90 1.703703704
64.77272727 4 95 1.614035088
66.38392089 5 97.36308397 1.574861098
64.77272727 6 95 1.614035088
61.36363636 7 90 1.703703704
57.95454545 8 85 1.803921569
55 9 80.66666667 1.900826446
total time required was equal to 15.61983471 seconds.
total distance covered was 1380 feet
total distance covered per segment was 153.33333333 feet.
average speed in feet per second for 1380 feet was 88.34920635
average speed in miles per hour for 1380 feet was 60.23809524
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