What is the Relative Velocity of Two Cars in Different Directions?

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The discussion centers on calculating the relative velocity of a Volkswagen Beetle and a Ford Mustang traveling in different directions. The Beetle moves east at 5.5 m/s, while the Mustang travels at 7 m/s at 75 degrees north of east. Participants clarify the use of vector components to find the Mustang's velocity relative to the Beetle, with calculations leading to a relative velocity of approximately -3.688i + 6.761j m/s. Confusion arises over the correct application of vector addition and subtraction, with users helping to identify calculation errors. The conversation highlights the importance of accurate vector resolution in solving relative motion problems.
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I need some help with this question:

Two cars, a Volkswagen Beetle travels east @ 5.5 m/s and a Ford Mustang traveling @ 75 degrees north of east @ 7 m/s. Both cars start from the same position and t=0.

a. What is the velocity of the Mustang with respect to the Beetle?
b. What is t when both are 60 m away from each other?
c. What is d after t=5 sec?


I am having a tough time with relative motion. I've read that section a million times. I can solve it by just treating it as a simple vector problem, but the teacher wants using the formula v=v' + V0. So for a do I resolve each vector into its i and j components and add? I'm really confused.
 
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\vec{v}_{M} = \vec{v}_{B} + \vec{v}_{M,B}, where M stands for Mustang, B for Beetle, and M, B for Mustang relative to Beetle. You know the velocity vectors, since the magnitudes and directions are given. Try to start with that.
 
\vec{v}_{M,B}=8.12 m/s @ 124 degrees

Is it now just a kinematics problem for b and c?
 
fsm said:
\vec{v}_{M,B}=8.12 m/s @ 124 degrees

Is it now just a kinematics problem for b and c?

How did you get that result? According to my calculations, this is wrong.
 
7*cos(75)i+7*sin(75)j=5.5i+0j+\vec{v}_{M,B}

-4.49i+6.76j=\vec{v}_{M,B}

R=sqrt((-4.49)^2+(6.76)^2)
R=8.12 m/s

theta=arctan(6.76/-4.49)
theta=-56.4 degrees
theta=180-56.4
theta=124 degrees
 
fsm said:
7*cos(75)i+7*sin(75)j=5.5i+0j+\vec{v}_{M,B}

Your calculation is wrong - the line above implies
\vec{v}_{M,B}=-3.688\vec{i}+6.761\vec{j}.
 
ok I think I found my error. Now I get 7.7 m/s @ 119 degrees.
 
Well I guess that one is wrong too. I don't see what I'm doing wrong.
 
\vec{v}_{M,B}=7.31\vec{i}+6.761\vec{j} is what I get now.
 
  • #10
Could anyone verify this?
 
  • #11
You got 7.31 by adding the i hats, rather than subtracting. It should be (sorry, no LaTeX)

(7*cos(75)-5.5)i+(7*sin(75))j=Vrelative
 
  • #12
Now when I did that radou said it was wrong. I have no idea now.
 
  • #13
fsm said:
Now when I did that radou said it was wrong. I have no idea now.

You didn't do that. You set the equation up correctly, and then miscalculated. EthanB is right, too.
 
  • #14
I still get -3.68i
 
  • #15
Please could someone tell me what I'm doing wrong? I don't get it.
 
Last edited:
  • #16
I'm not trying to be a pest, but anyone?
 
  • #17
radou said:
Your calculation is wrong - the line above implies
\vec{v}_{M,B}=-3.688\vec{i}+6.761\vec{j}.

fsm said:
I still get -3.68i

That's exactly what you should be getting. I think you misread what radou said: he said that the numbers you got were wrong, that you should've (from your data) gotten \vec{v}_{M,B}=-3.688\vec{i}+6.761\vec{j}.
 
  • #18
I get it now. I just have been working on this problem so much. Thank you both radou and EthanB for your help.
 
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