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Homework Statement:

Find the time that it takes for a car to reach the top of a hill.
The car can accelerate at 2.93 ft/s^2
The hill's slope = 6.4 %
The final velocity = 203mph
The initial velocity = 0mph
Homework Equations:

d = v^2/ 2a
t = v (f) / a
t = d (remaining) / v (f)
QUESTION:
If all the curves were removed, making a straight line from top to bottom, how quickly could a car reach the top of a mountain?
VARIABLES:
The mountain has a slope of 6.4%
The car's initial velocity is 0.
The car's top speed is 203 mph, (which will also be its final velocity as it crosses the finish line)
This car will accelerate at an average of .091 g (2.93 ft/s2).
The 12.42 mile long mountain course has over 156 turns. (Not sure what to do with this information. It says we remove all the turns, so the course should no longer be 12.42 miles, right?)
GIVEN EQUATIONS:
d=v^2/2a
t=v_f/a
t=d_remaining/v_f
I got that the time is equal to 1 minute and 41 seconds on one attempt, another gave 3 minutes and 7 seconds, and my latest attempt gave 4 minutes and 1 second. If anyone can help me get the answer, I would greatly appreciate it.
If all the curves were removed, making a straight line from top to bottom, how quickly could a car reach the top of a mountain?
VARIABLES:
The mountain has a slope of 6.4%
The car's initial velocity is 0.
The car's top speed is 203 mph, (which will also be its final velocity as it crosses the finish line)
This car will accelerate at an average of .091 g (2.93 ft/s2).
The 12.42 mile long mountain course has over 156 turns. (Not sure what to do with this information. It says we remove all the turns, so the course should no longer be 12.42 miles, right?)
GIVEN EQUATIONS:
d=v^2/2a
t=v_f/a
t=d_remaining/v_f
I got that the time is equal to 1 minute and 41 seconds on one attempt, another gave 3 minutes and 7 seconds, and my latest attempt gave 4 minutes and 1 second. If anyone can help me get the answer, I would greatly appreciate it.