Question

A person walks in the following pattern: 2.4 km north, then 1.9 km west, and finally 4.7 km south. (a) How far and (b) at what angle (measured counterclockwise from east) would a bird fly in a straight line from the same starting point to the same final point?

Answer 1

Answer:

a) Figure attached

b) For this case w have that A =(2.4 km)j, B= (-1.9 km) i , C= (-4.7 km)j

And the final position vector can be calculated adding the 3 vectors like this:

$s = A +B+C$

$s= (-1.9 km)i +(2.4 -4.7 km) j= (-1.9km)i + (-2.3 km)j$

We can find the magnitude of s like this:

$|s| = \sqrt{(-1.9)^2 +(-2.3)^2}=2.983$

And then we can find the angle with this formula:

$\theta = \tan^{-1} (\frac{-2.3 km}{-1.9 km})=50.44$

The other possibility is $\theta = 50.44+180 =230.44$

And since they want the angle measured from East the correct angle would be $\theta = 230.44$

Step-by-step explanation:

Part a

On the figure attached we have the vectors for the pattern described.

Part b

For this case w have that A =(2.4 km)j, B= (-1.9 km) i , C= (-4.7 km)j

And the final position vector can be calculated adding the 3 vectors like this:

$s = A +B+C$

$s= (-1.9 km)i +(2.4 -4.7 km) j= (-1.9km)i + (-2.3 km)j$

We can find the magnitude of s like this:

$|s| = \sqrt{(-1.9)^2 +(-2.3)^2}=2.983$

And then we can find the angle with this formula:

$\theta = \tan^{-1} (\frac{-2.3 km}{-1.9 km})=50.44$

The other possibility is $\theta = 50.44+180 =230.44$

And since they want the angle measured from East the correct angle would be $\theta = 230.44$