Answer:
d = 136.7 ft
Step-by-step explanation:
Because the truck move with uniformly accelerated movement we apply the following formula:
vf²=v₀²+2*a*d Formula (1)
Where:
d:displacement in meters (ft)
v₀: initial speed in ft/s
vf: final speed in ft/s
a: acceleration in ft/s²
Data
v₀ = 44.0 mi/h
1milla = 5280 ft
1h = 3600 s
v₀ = 44*(5280 ft) / (3600 s) = 64.5 ft/s
vf = 0
d = 47.0 ft
Calculation of the acceleration of the truck
We replace data in the formula (1) :
vf²=v₀²+2*a*d
0 = (64.5)²+2*a*(47)
-(64.5)² = (94)*a
a = -(64.5)² / 94
a = - 44.26 ft/s²
The acceleration (a) it's negative (-) because the truck is braking
Calculation of the minimum stopping distance of the truck to v₀ = 75.0 mi/h
v₀ = 75 mi/h = 75* (5280 ft) / (3600s) = 110 ft/s
We replace v₀ = 110 ft/s and a = - 44.26 ft/s² in the formula (1):
vf²=v₀²+2*a*d
0 = (110)²+2*(-44.26)*d
88.52*d = (110)²
d = (110)² / (88.52)
d = 136.7 ft