Question

A girl delivering newspapers covers her route by traveling 3.50 blocks west, 3.50 blocks north, and then 9.50 blocks east.

(a) What is her resultant displacement?
magnitude
blocks
direction
° north of east


(b) What is the total distance she travels?

blocks

Answer 1

(a) 6.95 blocks at
`30.2^(\circ)` north of east

Let's calculate the net displacement along the west-east direction first. Taking east as positive direction and west as negative direction,

`d_x = 9.50 - 3.50 = 6.00` blocks east

While displacement in north-south direction is

`d_y = 3.50` blocks north

So the magnitude of the resultant displacement is

`d=√(d_x^2 +d_y^2)=√((6.00)^2+(3.50)^2)=6.95` blocks

And the direction is

`\theta = tan^(-1) ((d_y)/(d_x))=tan^(-1) ((3.50)/(6.00))=30.2^(\circ)` north of east

(b) 16.50 blocks

The total distance travelled is simply the sum of the length of each path. We have:

3.50 blocks in the first part

3.50 blocks in the second part

9.50 blocks in the last part

So, the total distance travelled is

d = 3.50 + 3.50 + 9.50 = 16.50 blocks