Question on Velocity, two cars moving away from each other

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The discussion revolves around calculating the relative velocity of two cars moving away from each other, with specific attention to the horizontal and vertical components of their velocities. The calculations show that the horizontal component of car A's velocity should be negative, indicating motion in the opposite direction, which affects the overall velocity direction. The final velocity magnitude is calculated correctly, but there is confusion regarding the interpretation of the negative x component and its impact on the angle of motion. Additionally, another participant raises a similar question involving different velocities, indicating a broader uncertainty about the calculations. The thread emphasizes the importance of correctly identifying the direction of velocities in vector calculations.
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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