Question on Velocity, two cars moving away from each other

[Mrnickles]

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

 

[Doc Al]

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.

 

[tcsherpa]

Looks like you did it right.

 

[Mrnickles]

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

 

[mratwi36]

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.