Speed of Body After 2.06s with -3.93 m/s2 Acceleration

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Homework Help Overview

The problem involves calculating the speed of a body after a specific time under uniform acceleration. The initial speed is given as 4.37 m/s, and the acceleration is -3.93 m/s² over a duration of 2.06 seconds. Participants are discussing the implications of the negative acceleration and the interpretation of speed versus velocity.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the use of the kinematic equation for calculating final speed and express confusion over the results they obtain. There are questions about significant figures and the interpretation of speed as a scalar quantity versus velocity as a vector.

Discussion Status

There is an ongoing exploration of the problem with various interpretations being discussed. Some participants have suggested different numerical answers based on rounding conventions and the distinction between speed and velocity. The discussion reflects uncertainty about the correct answer and the potential for misinterpretation of the problem statement.

Contextual Notes

Participants mention that this is an assignment from a college online course, which may impose specific constraints or expectations regarding the format of the answer. There is also a concern about the accuracy of the answer expected by the submission system.

krazykaci
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The initial speed of a body is 4.37 m/s. What is the speed after 2.06s if it accelerates uniformly at -3.93 m/s^2. Answer in units of m/s.

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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What is the answer?
What did you get?
 
-3.7258 is what i got.
 
You are undoubtfully correct. Just be careful with significant figures: the solution is -3.73 m/s.

Why do you claim you are wrong?
 
its an assignment through a college online, and when i submit the answer -3.7258 or -3.73 it says I am wrong.
 
Try -3.72 m/s, then.
 
nope ... that didnt work either. I don't know what I am doing wrong? I am sure I am right.
 
PPonte said:
You are undoubtfully correct. Just be careful with significant figures: the solution is -3.73 m/s.

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.
 
It gets a little trickier because it asks "What is the speed", not "What is the velocity". Then it gives initial speed of 4.37 m/s, but since it is a speed, it implies no direction. Then it gives acceleration with a minus sign, implying direction, but not giving you a reference axis. So you don't know what direction relative to the initial speed it is accelerating.

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
loom91 said:
...even numbers don't get rounded up when the next digit is 5.
Are you sure about that? The digit after the 5 is an 8. I imagine 3.7258 rounds to 3.73.
 
  • #11
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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