Question

A disoriented physics professor drives 3.30 kmkm north, then 2.40 kmkm west, and then 1.70 kmkm south.

1- Find the magnitude of the resultant displacement, using the method of components.

2- Find the direction of the resultant displacement measured west of north.

3- Show that the resultant displacement found from your diagram is in qualitative agreement with the result you obtained by using the method of components.

From the method of components, Rx<0Rx<0 and Ry>0Ry>0, so R⃗ R→ is in the 2nd quadrant. This agrees with the vector addition diagram.

Answer 1

1) The magnitude of the resultant displacement, using the method of components is 2.88km

2)The direction of the resultant displacement measured west of north 33.7⁰ direction.

3)The magnitude of the resultant displacement of the professor is 2.88 km for both qualitative and components methods.

What is displacement?

Displacement is the change in an object's position, measured as a vector. It reflects both distance and direction from the starting point.

Given

Distances covered by the professor

3.3 km northward

2.40 km Westward

1.70 km Southward

Resolve in x and y components.

Dx = 2.40km

Dy = 3.30 -1.70 = 1.6 km

Resultant displacement

`R= \sqrt{ {Dx}^(2) + {Dy}^(2) }`

`R = \sqrt{ {2.4}^(2) + {1.6}^(2) }`

`R= √(5.76 + 2.56)`

`= √(8.32)`

= 2.88 km

The direction is found by

tanθ = Dy/Dx

= 1.6/2.4

= 0.6667

θ = tan⁻¹(0.6667)

= 33.7⁰

3) From the displacement diagram

AB is the distance north = 3.30 km

BC is the distance west = 2.4 km

CD is the distance south = 1.7 km

The displacement is distance DA

using Pythagorean theorem

ED = BC = 2.4km

AE = AB - CD

= 3.30 - 1.70 = 1.6 km

DA² = ED² + AE²

= 2.4² + 1.6²

DA = √8.32

= 2.88 km

Therefore, the displacement of the professor is the same qualitative and components methods.