# I must be wrong, instantaneous speed-Conceptual ?

1. Feb 22, 2012

### CherryXBOMB

1. The problem statement, all variables and given/known data
Before charging into my physics teachers office to demand an explanation for my being correct and yet being marked wrong I figured I should try to back up my argument.
Kim takes a trip leaving at noon and arriving at 4 pm. Here is the breakdown of her trip: 18 m/s for 3/4 hour, 27 m/s for 3 hours and 11 m/s for 1/4 hour.
A) What was her speed at 3 pm?

2. Relevant equations
avg speed=distance/time? i guess?

3. The attempt at a solution
I have typed everything out exactly as it appeared in my exam (which has been handed back to us, and will not be replicated). I wrote "not enough information" because to the best of my knowledge that questions is asking for instantaneous speed not the average speed. It also doesn't distinctively say, "the trip starts at 18 m/s for 3/4 hrs" etc. etc. maybe she was riding a bus that slowed down and sped up in increments? I don't know..please help! Maybe I'm just misunderstanding the concepts?

2. Feb 22, 2012

### cepheid

Staff Emeritus
If speed is constant, then instantaneous speed = average speed.

In this case, you have 3 different time periods during each of which Kim was travelling at a constant speed.

3. Feb 22, 2012

### Staff: Mentor

I think it's clear enough that it meant that the speed (instantaneous) was 18 m/s for 3/4 hours, etc. No way to dance around that! No reason to think average speed was meant.

4. Feb 22, 2012

### CherryXBOMB

Oh and just in addition: we specifically addressed this issue in discussion and she said using "not enough information" was a completely legitimate answer...I just don't understand how it could be wrong.

5. Feb 22, 2012

### Staff: Mentor

Did she say that "not enough information" was a legitimate answer in general (if you can back it up) or in regards to this particular problem?

The problem statement seems clear enough to me. I would not accept such an answer.

6. Feb 22, 2012

### cepheid

Staff Emeritus
I really think that that what was meant was that during the first 3/4 of an hour, the speed was constant, and equal to 18 m/s. As I pointed out before, if the speed was constant, then the instantaneous speed at any moment during this 3/4 period will be equal to the average speed (averaged over that 3/4 hour period). Both of them are 18 m/s.

During the next interval of 3 hours, the speed is constant at 27 m/s, so any time during this time interval, the instantaneous speed is 27 m/s. Furthemore, the average speed over this time interval is 27 m/s.

Etc.