Apickup truck traveling at a speed of 44.0 mi/h needs a minimum of 47.0 ft to stop. if the same truck is traveling 75.0 mi/h, determine its minimum stopping distance (in ft), assuming the same rate of acceleration.

d = 136.7 ft

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

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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