**a dog searching for a bone walks 6.5m south, then 8.2m at an angle of 31 degree north and east, and finally 15m west. find the dogs resultant displacement vector?**

**Answer:**

Consider the x-y axes in 2 dimensions.

Let +ve x direction be east, +ve y direction north, -ve x direction west and -ve y direction south.

Consider the dog to be starting from the origin of the coordinate system i.e. from (0,0) and travelling south

i.e. in the -ve y direction 6.5m. The new coordinates of the dog

are (x1, y1) = (0,-6.5).

Now the dog travels 8.2m at an angle 31∘ north-east i.e.

it begins to travel in the 4th quadrant at an angle 59∘ (= 90-31)

to the +ve x direction.

Let (x2, y2 ) be the point the dog reches at the end of 8.2 m.

Then x2-x1 = 8.2*cos(59∘)
and y2-y1 = 8.2*sin(59∘)

Thus (x2, y2) = (4.2233, 0.5288)

Now the dog travels 15m west i.e in the -ve x direction. So the y coordinate does not change and x3 = x2 - 15.

Hence (x3, y3) = (-10.777, 0.5288) which is the final position.

The displacement vector is the vector pointing from (0, 0) to (-10.777, 0.5288)

Hence the displacement of the dog is 10.79m (magnitude) from the origin

at an angle 2.81∘ clockwise from the -ve x direction (direction).