Question

A dog searching for a bone walks 3.50 m south, then 8.20 m at an angle of 30.0° north of east, and finally 15.0 m west. Use graphical techniques to find the dog’s resultant displacement vector

Answer 1

Steps to draw the graph-

Step 1: Firstly, we draw both of the axes.

Step 2: We then draw a vector of length 3.5 m (theoretical) along the (-x) axis.  

Step 3: Next, we draw a vector of length 8.2 m from the endpoint towards to top, at an angle of 60° from that point (N.B: 90-30, since it is required to make a 30°∠ with the current vertical; eventually the triangle being formed will be 90° minus the 30°.) Most importantly, the 8.2 m vector should be drawn in such a way so that it is higher than the starting point of the 3.5 m vector.  )

Step 4: Lastly, we draw a 15 m vector from the endpoint of the 8.2 m vector, running above the original vector, and to the left. This one should be drawn on the left side, and must be situated far enough.

∴ Triangle 1 ($t_{1}$) is of angles 30°, 60° and 90°;

On solving for the vertical side (the slope similar to the 3.5 vector), we get

(8.2/2) m = 4.1 m

Next, we solve for the opposite side via 4.1 × √3= 7.1 m

Moving on to Triangle 2 ($t_{2}$),

top side = (15 - 7.1) m = 7.9 m

similarly, right side = (4.1 - 3.5) m = 0.6 m

∴ last side = resultant displacement:

⇒ $\sqrt{(7.9)^{2}$ + (0.6)² = 7.9 m

and, direction of displacement = arcTan(y/x)

= arcTan(0.6/7.9)

= 4.3° north of west

Ans) The dog's resultant displacement vector is 7.9m, via 4.3° in the northwest direction.