They may be changed because they may find evidence of some thing that will change their perspective on things.
Answer: Option B: 1.3×10⁵ W
Explanation:


Work Done, 
Where s is displacement in the direction of force and F is force.

where, v is the velocity.
It is given that, F = 5.75 × 10³N
v = 22 m/s
P = 5.75 × 10³N×22 m/s = 126.5 × 10³ W ≈1.3×10⁵W
Thus, the correct option is B
Recall that

where
and
are the initial and final velocities, respecitvely;
is the acceleration; and
is the change in position.
So we have


(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)