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

double the distance = divide strength by 4

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

                                              ⇒      

here   k is the proportionality constant whose value is 1.

                                                            [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.