- #1

- 28

- 0

would I not use the kinamatic equ. v=v.+a(t) so v=4.37+-3.93(2.06)? cause when I do that I get the wrong solution?

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- Thread starter krazykaci
- Start date

- #1

- 28

- 0

would I not use the kinamatic equ. v=v.+a(t) so v=4.37+-3.93(2.06)? cause when I do that I get the wrong solution?

- #2

- 76

- 0

What is the answer?

What did you get?

What did you get?

- #3

- 28

- 0

-3.7258 is what i got.

- #4

PPonte

Why do you claim you are wrong?

- #5

- 28

- 0

- #6

PPonte

Try -3.72 m/s, then.

- #7

- 28

- 0

nope ....... that didnt work either. I dont know what im doing wrong? Im sure im right.

- #8

- 399

- 0

PPonte said:

Why do you claim you are wrong?

I think it's -3.72 m/s, even numbers don't get rounded up when the next digit is 5.

In any case, could you give us the exact statement of the problem? Also check that you are not making a mistake in reading the problem. Finally, there's the possibility that the answer known by the computer is wrong.

- #9

tony873004

Science Advisor

Gold Member

- 1,751

- 143

Speed is a scalar, not a vector, and should be a positive number.

I'd be tempted to try 3.73 m/s. (no negative sign).

- #10

tony873004

Science Advisor

Gold Member

- 1,751

- 143

Are you sure about that? The digit after the 5 is an 8. I imagine 3.7258 rounds to 3.73.loom91 said:...even numbers don't get rounded up when the next digit is 5.

- #11

- 399

- 0

tony873004 said:Are you sure about that? The digit after the 5 is an 8. I imagine 3.7258 rounds to 3.73.

I don't know about professional conventions, but I was taught that when rounding to the nth digit one should only consider upto the n+1)th digit, ignoring later digits. By this convention, you wouldn't round 5 to 6, you would treat it as a 5.

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