In an attempt to deliver a parcel on time, the dispatch rider had to ride 10 kilometers 15 degrees south east ,he then rode 11 kilometers 30 degrees north east and then takes a shortcut at 22 kilometers Find the rider's displacement

D = 4.05 km

Explanation:

In this case, it's really neccesary to analyze the given data.

First, the rider goes 10 km SE 15°. Then 11 km 30° to the NE direction. Then, he dispatch the order, and then, takes a shorcut to get back at 22 km heading W (This is the missing data of your exercise).

According to all these data, we have the magnitude of the displacement in each part of the travel, but to know the displacement we need to know the x and y components of these magnitudes, so we can really know the total displacement of the rider.

Let's cut this by parts. Part A would be the 10 km SE, Part B 11 km 30° NE and finally Part C would be the remaining 22 km

Part A:

in this case, is heading south east, so it's going in the x-axis positive but y-axis negative so:

Dxa = 10 cos15° = 9.659 km

Dya = 10 sin15° = 2.588 km (But as we are heading south is negative) = -2.588 km

Part B:

Heading NE, it's positive in the x-axis and y-axis so:

Dxb = 11 cos30° = 9.526 km

Dyb = 11 sin30° = 5.5 km

Part C:

Finally in this part is heading west, which means is heading in the x-axis negative, and it does not have a y component so:

Dxc = Dc = -22 km

Now that we have the displacement for each part and it's components, let's calculate the displacement in X and Y:

Dx = 9.659 + 9.526 - 22 = -2.815 km

Dy = -2.588 + 5.5 = 2.912 km

To calculate the magnitude of the displacement:

D = √Dx² + Dy²

D = √(-2.815)² + (2.912)²

Hope this helps

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