A boy runs 40m towards east. he then walks 30m towards north. The boy again runs 50m towards north making an angle of 30° with east. What is his displacement?

The displacement is 99.82 meters.

Explanation:

To use a standard notation, we can define:

North as the positive y-axis.

East as the positive x-axis.

If the initial position of the boy is (0, 0)

Then:

"runs 40m towars the east"

The new position is (0, 0) + (40m, 0) = (40m, 0)

"he then walks 30m towards North"

The new position is: (40m, 0) + (0, 30m) = (40m, 30m)

"The boy again runs 50m towards north making an angle of 30° with East."

By making a triangle rectangle, we can find the x and y components for this motion.

Y-component = 50m*cos(30°) = 43.3m

X-component = 50m*sin(30°) = 25m

Then the new position will be:

(40m, 30m) + (43.3m, 25m) = (83.3m, 55m)

Now we want to find the total displacement, this will be equal to the distance between the final position and the initial position.

The distance between two points (ax, ay) and (bx, by) is:

D = √( (ax - bx)^2 + (ay - by)^2)

In this case we want to find the distance between the points  (83.3m, 55m) and (0, 0), this will be:

D = √( (83.3m - 0)^2 + (55m - 0)^2) = 99.82 m

The displacement is 99.82 meters.

c.the charges if the proton and electron and gravity

not too sure thinking all accelerations are pointed down