Direction of resulting displacement

jamiebean · May 21, 2020

Printed textbook exercise 1.37: a physics professor drives distance s_1 north, then s_2 west, then s_3 south; asks (a) resultant displacement magnitude by components and (b) its direction.

Screenshot of a homework problem: displacement of 3.50 km north, 5.00 km west, and 1.30 km south. Part A answer is the square root of S2 squared plus (S1 minus S3) squared, equal to 5.46 km. Part B answer is acos of (S1 minus S3 over the square root of S2 squared plus (S1 minus S3) squared) times 180 over pi, equal to 66.3 degrees west of north.

I do not understand part b.
I know that x180/pi changes the value into a degree, but i do not understand what's going on inside the bracket. And how did the equation calculated the direction?

haruspex · May 21, 2020

jamiebean said:

Homework Statement:: attached below:
Relevant Equations:: displacement=final velocity-initial velocity

View attachment 263178 View attachment 263179

I do not understand part b.
I know that x180/pi changes the value into a degree, but i do not understand what's going on inside the bracket. And how did the equation calculated the direction?

Suppose the route goes from A north to B, west to C, then south to D.
Draw the diagram.
The angle you want is DAB. That is how much west of north the displacement is.
Now draw a line East from D until it hits the line AB at E. What is the distance AE? You already have the distance AD.
In terms of those, what is the cosine of DAB?

jamiebean · May 21, 2020

haruspex said:

Suppose the route goes from A north to B, west to C, then south to D.
Draw the diagram.
The angle you want is DAB. That is how much west of north the displacement is.
Now draw a line East from D until it hits the line AB at E. What is the distance AE? You already have the distance AD.
In terms of those, what is the cosine of DAB?

thank you! I understand now! :))

Direction of resulting displacement

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