Calculating Relative Velocity and Length Contraction in Special Relativity

[jlmccart03]

Homework Statement

The problem states: Racer A and Racer B have the same care length, but from a spectators view Racer A looks (1/2) that of Racer B. Also Racer B is traveling at a speed v = (c/2). I am to find the spedd of Racer A in the spectators frame of reference.

Homework Equations

Length contraction Δx = (Δx₀/ϒ)

The Attempt at a Solution

So what I did was simply realize that the length of the car stays the same in both the z and y directions, but changes in the x direction. This means that I don't have to worry about the z and y direction lengths as they are equal and cancel. From there I took the length contraction equation and got Δx_A = (Δx_B/2). The jist of that equation comes from the fact that x*y*z = ((x/2)*y*z) and everything but the x's cancel. That resulting answer looks like the length contraction eqaution. From there I just solved for v from the fact that ϒ=√(1-(v^2/c^2)). My final answer is v=(√3/4)c.

Now where I am confused is the fact that I never used the information of the c/2 for racer B. Was that unnecessary information given or did I do something wrong?

 

[jbriggs444]

Homework Statement

The problem states: Racer A and Racer B have the same care length, but from a spectators view Racer A looks (1/2) that of Racer B. Also Racer B is traveling at a speed v = (2/c). I am to find the spedd of Racer A in the spectators frame of reference.

Is that an accurate statement of the problem? A velocity of $\frac{2}{c}$ does not make sense. No velocity makes any sense unless a frame of reference is specified.

 

[jlmccart03]

Is that an accurate statement of the problem? A velocity of $\frac{2}{c}$ does not make sense. No velocity makes any sense unless a frame of reference is specified.

Sorry, edited it should be (c/2) not (2/c).

 

[phinds]

Also Racer B is traveling at a speed v = (2/c)

Check the units. 2/c is not a speed, it is the inverse of a speed

EDIT: OK, I see you caught that.

 

[jlmccart03]

Check the units. 2/c is not a speed, it is the inverse of a speed

EDIT: OK, I see you caught that.

Yeah, I typed it up wrong. It should be c/2.

 

[Praveen mahajan]

Homework Statement

The problem states: Racer A and Racer B have the same care length, but from a spectators view Racer A looks (1/2) that of Racer B. Also Racer B is traveling at a speed v = (c/2). I am to find the spedd of Racer A in the spectators frame of reference.

Homework Equations

Length contraction Δx = (Δx₀/ϒ)

The Attempt at a Solution

So what I did was simply realize that the length of the car stays the same in both the z and y directions, but changes in the x direction. This means that I don't have to worry about the z and y direction lengths as they are equal and cancel. From there I took the length contraction equation and got Δx_A = (Δx_B/2). The jist of that equation comes from the fact that x*y*z = ((x/2)*y*z) and everything but the x's cancel. That resulting answer looks like the length contraction eqaution. From there I just solved for v from the fact that ϒ=√(1-(v^2/c^2)). My final answer is v=(√3/4)c.

Now where I am confused is the fact that I never used the information of the c/2 for racer B. Was that unnecessary information given or did I do something wrong?

Yes you are right, though racer B is traveling and A is stationary. But for B it's stationary itsself but A is traveling and no need to use the speed given as due to speed , length is contracted

 

[jlmccart03]

Yes you are right, though racer B is traveling and A is stationary. But for B it's stationary itsself but A is traveling and no need to use the speed given as due to speed , length is contracted

Wait, now I am even more confused. So the problem asks for the speed of Racer A from the spectators view. A and B are both racing each other, but from the spectators viewpoint A is half the length of B. B is traveling at half the speed of light. Since we are focused on the racer A and not B then we don't have to worry about the speed v = c/2? I'm sorry, I am just really lost on how to look at these frames and whatnot.

 

[Praveen mahajan]

I hope if both are moving and in the same direction, then the concept of relative velocity is also need to adopt
Which is given as (v1-v2)/(1-v1v2/c^2)

 

[jlmccart03]

I hope if both are moving and in the same direction, then the concept of relative velocity is also need to adopt
Which is given as (v1-v2)/(1-v1v2/c^2)

So I have to take my 0.84c value and subtract it from the c/2 value? Or is that prior to finding my speed value of car A?

EDIT: According to your previous statement I took the values of v1=√(3/4)c and v2 = c/2. Using that equation I got a totale v = 0.64c. Would that sound correct?