Kevin walks 50.0 m to the east and 30.0 m at 30 degrees north of east. a) sketch kevin's move. b) what is the magnitude and direction of kevin's overall displacement?

a) see attached figure.

b) The magnitude of Kevin´s overall displacement is 77.5 m.

The direction of the overall displacement is 11.2° north of east.

Explanation:

Please, see the attached figure for a better understanding of the problem.

The resultant overall displacement will be the sum of the vectors A+B (see figure).

The vector A is horizontal, it has no vertical component. Then:

A = (50.0 m, 0)

The vector B, in change, has a vertical and horizontal component. Applying trigonometry of right triangles we can find the components of B:

By trigonometry:

cos θ = adjacent / hypotenuse

sin θ = opposite / hypotenuse

In this case:

cos 30° = x-component of B / magnitude of B

cos 30° = x-B / 30.0 m

30.0 m · cos 30° = x-B

x-B = 26.0 m

sin 30° = y-component of B / 30.0 m

30.0 m · sin 30° = y-B

y-B = 15.0 m

Then, the vector B will be:

B = (26.0 m, 15.0m)

The overall displacement will be:

R = A + B

R = (50.0 m, 0) + (26.0 m, 15.0 m)

R = (50.0 m + 26.0, 0 + 15.0 m)

R = (76 m, 15 m)

The magnitude of the overall displacement will be:

The magnitude will be 77.5 m.

Using trigonometry, we can find its direction (see figure):

cos θ = 76 m / magnitude of R

θ = 11.2°

The direction of the overall displacement is 11.2° north of east.

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach