Given the time, the final velocity and the acceleration, we can calculate the initial velocity using the kinematic equation A:
A skateboarder flies horizontally off a cement planter. After a time of 3 seconds (Δt), he lands with a final velocity (v) of −4.5 m/s. Assuming the acceleration is -9.8 m/s² (a), we can calculate the initial velocity of the skateboarder (v₀) using the kinematic equation A.
Given the time, the final velocity and the acceleration, we can calculate the initial velocity using the kinematic equation A:
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Answer:
W = ½ m v²
Explanation:
In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation
We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved
initial instant. before separation
p₀ = m v
final attempt. after separation
= m /2 0 + m /2 v_{f}
p₀ = p_{f}
m v = m /2 
v_{f}= 2 v
this is the speed of the second part of the ship
now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body
initial energy
K₀ = ½ m v²
final energy
= ½ m/2 0 + ½ m/2 v_{f}²
K_{f} = ¼ m (2v)²
K_{f} = m v²
the expression for work is
W = ΔK = K_{f} - K₀
W = m v² - ½ m v²
W = ½ m v²
