A student starts at position (4,4) m and undergoes three displacements: d1 = (2, -3) m, d2 = (-5, 0) m, and d3 = (1, 6) m.
What is the final position of the student? Show your solution graphically and numerically.
please help!!

The final position is  (2, 7)m

Explanation:

When we work with coordinate pairs, the addition works as:

(a, b) + (c, d) = (a + c, b + d)

So, for example, if we start at (a, b), and we have a displacement d = (1, 1)

we just need to solve:

(a, b) + (1, 1) = (a + 1, b + 1)

Now, in this case, we start at (4, 4)m

first, we have d1 = (2, -3) m

After this displacement, the position is:

(4, 4)m + (2, -3)m = (4 + 2, 4 - 3)m = (6, 1)m

Now we have a displacement d2 = (-5, 0) m

After this, the position is:

(6, 1)m + (-5, 0)m = (6 -5, 1 + 0)m = (1, 1)m

After this, we have the final displacement d3 = (1, 6) m, so the final position will be:

(1, 1)m + (1, 6)m = (1 + 1, 1 + 6)m = (2, 7)m

Below you can see a rough sketch of the path that the student take, where he/she starts at point A.

2b2t

explanation:

2b2t

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge