Question on Velocity, two cars moving away from each other

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Homework Help Overview

The discussion revolves around a problem involving the velocities of two cars moving away from each other, specifically focusing on their vector components and the resultant velocity. Participants are examining the calculations related to the horizontal and vertical components of velocity.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the relative velocity of one car with respect to another by breaking down their velocities into components. Some participants question the sign convention used for the velocities, particularly regarding the direction of motion. Others discuss the validity of the calculations and the implications of the angle derived from the components.

Discussion Status

There is an ongoing examination of the calculations presented, with some participants affirming the correctness of the approach while others suggest reconsidering the assumptions made about the direction of the velocities. Multiple interpretations of the results are being explored, particularly in relation to the angle derived from the velocity components.

Contextual Notes

Participants are working under the constraints of the problem as presented, with some expressing uncertainty about their calculations and seeking clarification on specific aspects of the velocity components and angles involved.

Mrnickles
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i would greatly appreciate any help towards this...i thought my workings were correct, but i don't think they are now...
http://i21.photobucket.com/albums/b283/NMchugh/Various/velocity.png Question


vAx = 80 vAy = 0 vBx = 50 cos(60 vBy = 50 sin(60

First we look at the horizontal X- component
vBAx = vBx - vAx
vBAx = (50 cos60) - (+80)
vBAx = 25 - 80 = -55km/h

Next we look at the vertical y- component
vBAy = vBy - vAy
vBAy = (50 sin60) - (0)
vBAy = 43.3km/h

vBA = √(vBAx)² + (vBAy)² = √(-55)² + (43.3)² = 70
 
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Looks OK to me except that you have motion in the negative x direction as positive. For example, I'd say that vAx = - 80 km/h, not +80. That will end up affecting your direction for the velocity.
 
Looks like you did it right.
 
my friend seems to think this is a check i can do to see if my answer is correct (gives the correct angle)
does this answer mean my workings are wrong?

tan∅ = vABy = 43.3 = 0.41, ∅ = tanˉ¹ ( 0.41) =22.4
vABx 105
 
i have the same question as well but a is 95 KM and b 125 KM,got the the angle point and gt stuck there thought my answer is wrong,so i don't know what to do.
 

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