Physics. One more problem..Thank you!?

L

Larisa

New member
The niminum distance required to stop a car moving at 38.0 mi/h is 47 ft. what is the minimum stopping distance for the same car moving at 71 mi/h, assuming the same tare of acceleration?
 
since the acceleration is constant,
u can use V(t)^2 - V(0)^2 = 2*a*s
case I: V(t) = 0, V(0) = 38mph, a, s=47ft
=> -(38)^2 = 2a*47
Case II : V(t) = 0, V(0) = 71mph, a, s=?
=> -(71)^2 = 2a*s
take ratio and cancel terms u will get
s= 164 ft.
 
1 mph = 0.44704 m / s
38 mph = 16.98752
71 mph = 31.73984
1 meter = 3.2808399 feet
47 f = 14.325599978225088033097866189691


Vf2 = Vi2 + 2AS

0 = (16.98752)2 + 2A(14.325599978225088033097866189691)
0 = 288.5758357504 + 2A(14.325599978225088033097866189691)
-288.5758357504 = 2A(14.325599978225088033097866189691)
-144.2879178752 = A(14.325599978225088033097866189691)
-10.072033150061263414468085106384 = A


0 = 31.739842 + 2(10.072033150061263414468085106384)S
0 = 1007.4174432256 + 2(10.072033150061263414468085106384)S
-1007.4174432256 = 2(-10.072033150061263414468085106384)S
-503.7087216128 = (-10.072033150061263414468085106384)S
50.010629840881349567068104890758 = S
The stopping distance at 71 mi/h is roughly 50 metres or 164 feet
 
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