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
To hit the target the crew drop the crate before the plane is directly over the target. It is because <span>because the cargo has forward velocity and therefore before it reaches the ground it travels some distance. The answer is A. Hope it helps. </span>
Explanation:
If velocity is increasing uniformly with time, this means the acceleration is constant. So we can use this formula,
Initial Velocity+Final Velocity/2.
Our total Velocity is 60 m/s. Let convert minutes to seconds, which is 120 seconds so we get
60m/s/120 s which we get 0.5 m/s.
Average velocity is 0.5 m/s^2.
the production of new machines as a result of processes using science and knowledge
The action and reaction force are equal and opposite