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**vector problem..PLEASE HELP!!!**

A car is driven 150 km west and then 30 km southwest. What is the displacement of the car from the point of origin (magnitude and direction)?

i have no clue how to start..could someone help me?

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- Thread starter confusedaboutphysics
- Start date

- #1

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A car is driven 150 km west and then 30 km southwest. What is the displacement of the car from the point of origin (magnitude and direction)?

i have no clue how to start..could someone help me?

Last edited:

- #2

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- #3

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Right, and I would assume southwest to mean to have a reference angle of 45 degrees.

- #4

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i know that V1x = -150 and V1y = 0. but i'm confused how to find V2x and V2y. do i use the Vx=V1x + V2x? but i dont know what Vx is. and i know that to find Vx you use Vx=Vcos(angle) but i dont have V either.

where do i go from here?

where do i go from here?

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- #5

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someone help please! i dont know where to go from here! (see above message)

- #6

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[itex] \vec{v_2} = 30km \ southwest [/itex]

Vector 1 is purely East West so it has no north south component.

Vector 2 has components in both north south and east west. SW is 45 degrees south of west. Break it up into its components.

- #7

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First of all, V(x) = V(x1) + V(x2) and V(y) = V(y1) + V(y2), right? It appears you know this already.

You've solved for V(x1) and V(y1). For V(x2) and V(y2), you must use your formulas. That is, V(y2) = V times the sine of the angle and V(x2) = V times the cosine of the angle. The angle is 45 and the magnitude is 30. Just plug in the values and get the x and y components of the second vector. Then plug these results into the formula displayed at the top of this post and you have the x and y components of the final vector. Using Pythagoras' Theorem and basic trigonometry, you can then find the magnitude and direction of the final vector. Do you understand?

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