Gravity's effect on velocity question

[lzh]

here is the problem:
"a first aid kit is dropped by a rock climber who is descending steadily at 2.2m/s"
after 2.9s, what is the velocity of the first aid kit?

the init. velocity of the kit is 2.2 like the climber, its accerlation is 9.8 which is the accerlaeration of gravity.
so:
final velocity=2.2+9.8(2.9)
which equals to 30.62
This problem really should be easy, but for some reason my web homework service tells me that the answer is wrong.
whats wrong?

 

[Hootenanny]

Your equation is not quite right. Have you seen any kinematic equations before?

 

[lzh]

but The final velocity is equal to the initial velocity plus the acceleration multiplied by the time.
and that's what my equation is

 

[rsk]

yes, I would have gone for v = u + at.

I can't see yet why this isn't right. Anyone?

 

[lzh]

Hootenany what exactly did you mean was not quite right?

 

[Hootenanny]

Sorry, my bad. I thought we were talking about distances not velocity, I didn't read the question correctly. Have you tried entering - 30.62? Does web assign specify a number of significant figures?

 

[lzh]

no, it accepts any answer to the 1% of the exact answer.
i tried 30.6392 which is more exact because i used 9.80665 instead of 9.8, and it was wrong

 

[Hootenanny]

Still have you tried entering -30.62?

 

[lzh]

no, i have not. But wouldn't 30.6392 be a even closer number?

 

[rsk]

i think the point is the negative sign, and it's worth a try, although the speed of the climber is given as positive.

 

[Hootenanny]

no, i have not. But wouldn't 30.6392 be a even closer number?

No, since your data is only accurate to two significant figures, anything beyond that is simply 'noise', i.e. meaningless. The most number of significant figures you should quote is three, i.e. one more than your input data.

 

[Hootenanny]

i think the point is the negative sign, and it's worth a try, although the speed of the climber is given as positive.

speed is always positive.

 

[lzh]

i don't know about that...
this question has two parts. the first part is what i posted and the second part asks for how far the kit is from the climber. I got the second part right with:
-i figured out the kit's displacement to be 47.589 so that minus the displacement of the climber is my answer, 41.209.

so i used that number in the kinematics forumlas that i knew to find the final velocity, and all of them resulted in 30.6...

 

[lzh]

heres the website's description:
a) Significant digits and precision
The computer carries out all calculations to at least six significant digits. Do not use "significant figures" algorithms to round off your answer. Do not round off 'intermediate' calculations. Six digits are shown in solutions.
To be scored as correct, an answer must be within 1% of the computer's answer (except for an answer of zero, which must be exact). You will be informed of any exceptions to this tolerance.
i'm using utexas's web homework page btw.

 

[Hootenanny]

If you are instructed not to round off then don't, enter the number as it is displayed on your calculator.

 

[rsk]

speed is always positive.

Oops of course.

But you know what I meant.

 

[lzh]

ok, i'll try 30.62

 

[lzh]

it was wrong...

 

[Hootenanny]

Oops of course.

But you know what I meant.

I know I was being pedantic but it is usually best to clarify.

it was wrong...

Have you tried -30.62 yet?

 

[lzh]

yeah, it worked... why?
edit: yeah speed is positive but velocity can be neg.
but i still don't see why

 

[rsk]

downwards direction taken as negative

 

[Hootenanny]

yeah, it worked... why?

Because it is convention to define vertically upwards at the positive direction. For example 30m/s would usually mean 30m/s upwards; -30m/s would usually mean 30m/s vertically downwards. However, the wording of the question didn't help.

 

[lzh]

yeah, it was kinda confusing.
well thanks for the help guys! appreciate it!