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the anwser is c is always the best anwser
Answer: 7.41 m/s
Explanation: By using the law of of energy, kinetic energy of the brick as it falls equals the potential energy before falling.
Kinetic energy = mv²/2, potential energy = mgh
mv²/2 = mgh
v²/2 = gh
v² = 2gh
v = √2gh
Where g = 9.8 m/s², h = 2.80m
v = √2×9.8×2.8 = 7.41 m/s
Let us consider body moves a distance S due to the force F.
Hence the work by the body W = FS
If the force is not along the direction of displacement,then the work by a body for travelling a distance S will be -
![W=[ Fcos\theta]*S](https://tex.z-dn.net/?f=W%3D%5B%20Fcos%5Ctheta%5D%2AS)
where 
is the component of the force along the direction of displacement.
As per the question the power P is given as -
Hence alternative definition of power P = F.V
Answer:
2.69 m/s
Explanation:
Hi!
First lets find the position of the train as a function of time as seen by the passenger when he arrives to the train station. For this state, the train is at a position x0 given by:
x0 = (1/2)(0.42m/s^2)*(6.4s)^2 = 8.6016 m
So, the position as a function of time is:
xT(t)=(1/2)(0.42m/s^2)t^2 + x0 = (1/2)(0.42m/s^2)t^2 + 8.6016 m
Now, if the passanger is moving at a constant velocity of V, his position as a fucntion of time is given by:
xP(t)=V*t
In order for the passenger to catch the train
xP(t)=xT(t)
(1/2)(0.42m/s^2)t^2 + 8.6016 m = V*t
To solve this equation for t we make use of the quadratic formula, which has real solutions whenever its determinat is grater than zero:
0≤ b^2-4*a*c = V^2 - 4 * ((1/2)(0.42m/s^2)) * 8.6016 m =V^2 - 7.22534(m/s)^2
This equation give us the minimum velocity the passenger must have in order to catch the train:
V^2 - 7.22534(m/s)^2 = 0
V^2 = 7.22534(m/s)^2
V = 2.6879 m/s
