0:00
Hello, everyone.
00:01
In this problem, we're asked to find the speed of a car, of a rail car, given that in that, in the reference frame of that car, and other car's velocity is minus 0 .2c, whose relative, whose velocity relative to the ground is measured to be 0 .5c.
00:21
So a stationary observer measures cart, let's say cart ones, speed to be 0 .5c.
00:30
Whereas an observer on cart 2 measures the speed of this cart to be minus 0 .2c.
00:38
So what we're asked, what we have to find is the speed of this cart 2, right, such that this criteria is satisfied.
00:48
So essentially what we know, i've already labeled this in kind of a suggestive manner, but the velocity or the speed of cart 1 in the original frame, the rest frame, so the grounds, rest frame is 0 .5c and its velocity in the moving frame of quartz 2 is minus 0 .2c.
01:09
So then all we have to do is really use this, you know, velocity of the long run's transformation where we see that the velocity of a moving object in a, in a moving frame is going to be u prime is equal to u minus v or 1 minus u times v over c squared.
01:30
And here we're told what u prime is and what u is, and all we have to do is rearrange this for v, and then plug in the values that we're given.
01:37
So doing so, first we multiply through this equation by 1 over or 1 minus u times u over c squared.
01:46
That gets us to this line over here...