Question on Velocity, two cars moving away from each other

  • Thread starter Mrnickles
  • Start date
  • #1
5
0
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
 

Answers and Replies

  • #2
Doc Al
Mentor
44,990
1,263
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.
 
  • #3
1
0
Looks like you did it right.
 
  • #4
5
0
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
 
  • #5
8
0
i have the same question aswell 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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