In order to be considered a vector, a quantity must include Magnitude (A) and Direction (D).
The acceleration of the electron is larger than the acceleration of the proton.
The reason for this is that the mass of the electron is smaller (about 1000 times smaller) than the mass of the proton. The two particles have same charge (e), so they experience the same force under the same electric field E:
However, according to Newton's second law, the force is the product between the mass particle, m, and its acceleration, a:
which can be rewritten as

we said that the force exerted on the two particles, F, is the same, while the mass of the electron is smaller: therefore, from the last formula we see that the acceleration of the electron will be larger than that of the proton.
Answer:
130 km at 35.38 degrees north of east
Explanation:
Suppose the HQ is at the origin (x = 0, y = 0)
So the coordinates of the helicopter after the 1st flight is


After the 2nd flight its coordinate would be:


So in order to fly back to its HQ it must fly a distance and direction of
north of east
In a velocity-time graph, the area under the curve represents the distance.
The distance traveled from 10s to 18 s is

Final Answer: The distance traveled is 40 m from time 10 s to 18 s.