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Answer:
The car overtakes the truck at a distance d = 3266.2ft from the starting point
Explanation:
Problem Analysis
When car catches truck:
dc = dt = d
dc: car displacement
dt: truck displacement
tc = tt = t
tc: car time
tt : truck time
car kinematics :
car moves with uniformly accelerated movement:
d = vi*t + (1/2)a*t²
vi = 0 : initial speed
d = (1/2)*a*t² Equation (1)
Truck kinematics:
Truck moves with constant speed:
d = v*t Equation (2)
Data
We know that the acceleration of the car is 3.00 ft / s² and the speed of the truck is 70.0 ft / s .
Development problem
Since the distance traveled by the car is equal to the distance traveled by the truck and the time elapsed is the same for both, then we equate equations (1 ) and (2)
Equation (1) = Equation (2)
(1/2)*a*t² = v*t
(1/2)*3*t² = 70*t (We divide both sides by t)
1.5*t = 70
t = 70 ÷ 1.5
t = 46.66 s
We replace t = 46.66 s in equation (2) to calculate d:
d = 70*46.66 = 3266.2ft
d = 3266.2 ft
Answer: I = 3.6 m3
(C)
Explanation:
moment of inertia for spherically shaped object around it's center is given as
I = (2/5) mr²
substituting the r = 3m²
I = (2/5)*(9) m3
I = 3.6 m3
