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# Introductory Physics Help

[From: ] [author: ] [Date: 11-10-07] [Hit: ]
so the faster speed is 56/30 greater than the original speed,(56/30)^2 = 3.that means the stopping distance at the higher speed is 3.30 miles / hr = 30 * 5250 /3600 = 43.a = -43.75^2/(2*48) = -19.......
The minimum distance required to stop a car moving at 30.0 mi/h is 48.0 ft. What is the minimum stopping distance for the same car moving at 56.0 mi/h, assuming the same rate of acceleration?

Thank you so much for the help.
i am really trying to understand the concepts?

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vf=final velocity = 0
v0=initial velocity
a=acceleration
d=stopping distance

solving for d, we have

d= v0^2/2a

(since a is negative)

we see from this that if you hold the acceleration constant, the stopping distance varies as the square of the initial velocity

so the faster speed is 56/30 greater than the original speed, so the stopping distance will be

(56/30)^2 = 3.48 times greater

that means the stopping distance at the higher speed is 3.48 x 48 ft = 167ft

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equation of concern is 2ad=v2^2 - v1^2
v2=0 so equation becomes
solve for a
a =-v1^2/2d
if we want to answer in feet need to convert speed from mph to ft/s
30 miles / hr = 30 * 5250 /3600 = 43.75 ft / s
a = -43.75^2/(2*48) = -19.94 ft / s^2
now use same equation but solve for d
change 56 mph to ft/s
56*5250/3600=81.67ft/s
d=-v1^2/2a = -81.67^2/(2*-19.94)
d=167.2 ft
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