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eflickinger
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Homework Statement
A vehicle weighs 3000 lbf and has an initial velocity of 68 mph. If the braking system is designed to stop the vehicle at constant power, calculate the stopping distance in feet, if the stopping power is 100 hp.
m = 3000 lbf / 32.2 ft/s2 = 93.17 (lbf-s2)/ft
v = 68 mph = 68 miles/hr x 5280 ft/mile x 1hr/3600s = 99.73 ft/s
P = 100 hp = 100 hp x 550 (ft-lbf)/s / 1 hp = 55,000 (ft-lbf)/s
Homework Equations
Fb = Braking Force
P = Energy / Time
Energy = 0.5*m*v2
Time (to stop) = t = (mv)/Fb
Distance (to stop) = x = 0.5*(m/Fb)*v2
The Attempt at a Solution
We need to calculate x, the stopping distance.
m and v are defined, however Fb is not.
We can find the time (to stop) based on the power = energy / time equation since we have both energy and power.
Therefore, Fb = (mv)/t
Plugging back into distance (x),
0.5*(m/((mv)/t))*v2
This makes x, the stopping distance = 0.5*t*v
t = Power / Energy
Energy = 0.5*m*v2
Energy = 0.5(93.17)(99.73)2 = 463,337.81 ft-lbf
Power = 55,000 (ft-lbf)/s <-- As stated previously
Therefore, t = 463,337.81 / 55,000 = 8.42 seconds
Now plugging back into the stopping distance formula we derived:
0.5(8.42s)(99.73ft/s) = 420.08 ft
But somehow this isn't the correct answer according to my online homework (which also doesn't supply the correct answer)?
Thank You,
Evan