What is the final displacement of the car relative to its starting point?

Question

A vehicle travels 25 km at an angle of 20 degrees north of east, then turns and travels a distance 35 km of at an angle of 50 degrees west of north and finally turns and travels a distance of 10 km at an angle of 10 degrees south of west.

Answer 1

with a diagram I get 32 km at an angle 66 degrees north of west

Answer 2

Thuto,

 

First let's specify all angles with respect to the x-axis and determine qualitatively the directions of travel with respect to the x and y axes.  It helps to actually graph this as you go and gives you a check on your final answer!

 

The first direction is already measured with respect to (wrt) the x-axis so we can leave that alone.

The second direction is 50° N or W which is 40° W of N.

The third direction is already measured wrt the x-axis.

 

Next write the 3 directions in terms of their west i components (x-direction) and j components (y-direction).

Bold letters indicate vector notation.

 

X₁ = 25·(cos 20° i + sin 20° j) km = (23.5 i + 8.55 j) km

x₂ = 35·(-cos 40° i + sin 40° j) km = (-26.8 i + 22.5 j) km

x₃ = 10·(-cos 10° i - sin 10° j) km = (-9.85 i - 1.74 j ) km

 

The (-) signs indicate that the directions translate into motion to the left (-x) or down (-y)

 

Now add all the x-components and y-components separately to get the final position.

 

x_f = (23.5-26.8-9.85) i + (8.55+22.5-1.74) j km

    = -13.2 i + 29.3 j km (final position vector)

 

The displacement is [(-13.2)² + (29.3)²]^½ = 32 km at an angle of 66° North of West

 

The last comes from using the x and y-components of the final displacement vector to find tan θ