Answer:
Technician A is correct
Explanation:
As per the question:
We know that kinetic energy of an object is given by:
[tex]KE = \frac{1}{2}mv^{2}[/tex]
where
v = speed of the object
Now, let us double the speed:
v' = 2v
Now,
[tex]KE' = \frac{1}{2}mv'^{2}[/tex]
[tex]\frac{1}{2}m(2v)^{2}[/tex]
[tex]KE' = 4.\frac{1}{2}mv^{2} = 4KE[/tex]
Thus the Kinetic energy becomes 4 times of its initial value.
