What is the distance and speed of a girl on a bicycle passing a bus stop?

[RichardT]

Homework Statement

A girl is riding a bicycle along a straight road at constant speed, and passes a friend standing at a bus
stop (event #1). At a time of 60.0 s later, the friend catches a bus (event #2).
(a) In the frame of reference of the earth, what is the distance separating these events?
(b) If the distance separating the events is 126 m in the frame of the girl on the bicycle, what is the
bicycle's speed?

Homework Equations

v = d/t

The Attempt at a Solution

For part a. d = vt where delta t (time) equals 60 seconds and I just left velocity as its variable so d = 60(v) so d = 60v

Part B where delta distance = 126m. This is where i start to run into problems. At first i thought the answer would be found using just v = d/t and what i did was v = 126m/60s which equals 2.1m/s. However, i think its incorrect because it asks for the velocity in the frame of reference of the girl on the bicycle. If anyone could clear this up it would be great. CORRECTION: it actually says the DISTANCE separating the events is 126m in the FRAME OF THE GIRL ON THE BYCYCLE. Would This still change my results?

 

[Mindscrape]

Try using the space time invariant :)
-c^2t^2=x'^2-c^2t'^2

Alternatively, you could take your approach. However, you must note that time is not the same in the girl's frame. Her time is shorter than 60 seconds.

 

[RichardT]

Wait, so the answer is not 2.1 m/s? We never learned the equation and i think time dilation is what your talking about but my teacher said not to use time dilation.

 

[Mindscrape]

You're not going to get a very good numerical answer without a binomial, or taylor, expansion. To some extent, yes, the "answer" is 2.1m/s, but only because the whole point of the problem is that the bicycle is going so slow that relativistic effects don't come into play for the first 16 digits (more than your calculator or computer program will give you). Maybe your teacher said not to use time dilation because he was afraid of people doing plug and chug. If you're not allowed to use time dilation, try the space time invariant, you'll like it.

 

[RichardT]

Right Now I am thinking something different. The question states that event #1 is when the girl passes a friend and event # 2 is when the friend gets on the bus. Therefore since the friend did not actually MOVE the distance separating the events is 0 m.

For Part B, I am starting to believe it does not include any equations. In the second part the friend is moving away from the bicycle at the same speed as the bicycle is moving (but in opposite direction) and relative to the bicycle the two events are indeed separated by d = vt . The "v" I calculated is the velocity of the friend as seen relative to the bike. But it is also the speed of the bike relative to the earth, which I believe is what they want. The speed of the bike relative to the girl riding it is zero. The question I am not sure is are they asking for the speed of the bike relative to the girl which would be zero or the speed of the bike relative to Earth which will be 2.1m/s. Anyone want to check if this is correct?

 

[Mindscrape]

You've only got part of the picture down. You're right, the friend doesn't move, meaning that the friend is only "moving" along in time. The spatial distance separating the events in the friend frame is 0, absolutely. You're also right, the velocity of the bike is the same as the velocity that the friend sees the girl moving at. However, you are completely wrong if you think the two times are equivalent. Moving clocks tick slower, it's one of the huge revelations of special relativity.

The way to see this problem is essentially that the friend moves along in time, and the girl moves along in space and time. Have you been talking about the "length" or "invariance" of space-time in class? If not, I see no other way to view this problem if you're not supposed to be viewing it as time dilation.

 

[RichardT]

No, we haven't learned any of the topics you mentioned. Also this question is in the textbook and the question comes before the unit of time dilation. So i do not think that a question would be located in a textbook before the topic is even introduced to. However, I see what you are trying to say and I believe that you are correct but what other options are there than when i haven't learned any of these topics? Since the velocity of the bike in the frame of reference to the Earth or the friend is what i thought to be was 2.1m/s and you propose a reason why its not 2.1m/s, should i think that the question is asking what is the speed of the bike relative to the GIRL which would be zero? Also for Part A, would the difference separating the two events in the FRAME OF THE EARTH be zero as well?

 

[Mindscrape]

The Earth frame and friend frame are the same frame, just as the girl frame and the bike frame are the same frame. You can get the velocity of the moving frame (i.e. the velocity of the bicycle) by saying v=d'/t' or v=d/t. The problem is that you are given d' and t, and in order to get your 2.1m/s you said v=d'/t, which is not valid.

Let's just approach it from the time dilation perspective.

$$v=d'/t'=\frac{d'}{t \sqrt{1-v^2/c^2}}$$

You could potentially go through the algebra and solve for v. You'd get the same answer as using the space time invariant, which I guess has not been covered yet. Either way, you'll end up noticing that the difference in velocity between the non-relativistic (your solution) and the relativistic solution is to first order x'^2/(ct)^2. c is a really big number compared to these others, so the difference is not much. That is why your non-relativistic answer works, and it is why we don't see special relativity in our everyday lives.

 

[RichardT]

Why is time not t', i understand why the distance because it is in the frame of reference of the girl on the bike its d', but why is the time given, 60 seconds, just t and not t' as well?

 

[Mindscrape]

I don't really know how to answer that, why t is not also t', other than that's just the way the world works. At really low speeds, the speeds of our every day lives, they are essentially equivalent. I mean, if I get in my car and drive for 5 minutes to a friend's house, we don't say hey, "I'll be there in 5 minutes, my frame's time." But if I get in a rocket and fly for 2 hours to my friend's planet, we better make sure who's time we are talking about.