Calculating Displacement: Magnitude & Direction of a Car's Motion

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Homework Help Overview

The problem involves calculating the displacement of a car that travels in two segments: first north and then east. The original poster attempts to find both the magnitude and direction of the final displacement using basic kinematic principles.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations for magnitude and direction, with one participant noting a potential misunderstanding regarding the angle's reference direction.

Discussion Status

The discussion is ongoing, with participants exploring the implications of directional notation in trigonometric calculations. There is an indication that a drawing may help clarify the misunderstanding regarding the angle measurement.

Contextual Notes

There appears to be confusion about the correct interpretation of directional angles, specifically the distinction between "North of East" and "East of North." This may affect the understanding of the problem setup.

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Homework Statement


A car travels north with a velocity of 5.2-m/s for 92-seconds. It then turns east, and with the same speed travels for a further 22-seconds. Calculate the magnitude of the final displacement of the car.

Now give the direction of the final displacement of the car. Give your answer in degrees North of East.

Homework Equations


pythargaros

The Attempt at a Solution


Hi i have calculated the magnitude (492m) by doing 5.2 x 92 and 5.2 x 22 and using pythagaros

For the 2nd part (direction in degrees). I did tan^-1(114.2/478.4) and got 13 degrees. this was wrong
 
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Hello Zbrah, welcome to PF :smile: !

13 degrees North of East ?
 
BvU said:
Hello Zbrah, welcome to PF :smile: !

13 degrees North of East ?
I wrote that in my answers, but it was wrong :(
 
Make a little drawing. You'll immediately see the difference between 13 degrees North of East and 13 degrees East of North...
 
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