Find the time when velocity = 10 m/s

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Homework Help Overview

The discussion revolves around determining the time when the velocity, given by the equation v = (5.4t - 4.4t^2)i + 8.8j, equals 10 m/s. Participants are exploring the implications of solving for time in this context.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • One participant attempts to solve the equation by squaring both sides to eliminate the radical, leading to a quadratic equation. They express concern about the highest power of t being up to 4 and question the correctness of their method.
  • Another participant notes that the roots of the quadratic equation are imaginary and questions the accuracy of the original problem statement.
  • Further discussion includes whether it is legitimate to move a negative number under the radical, with some participants exploring the implications of obtaining imaginary results in the context of time.

Discussion Status

The discussion is active, with participants questioning the validity of the original problem and the implications of obtaining imaginary solutions. There is no explicit consensus, but various interpretations and concerns regarding the physical meaning of the results are being explored.

Contextual Notes

Participants are grappling with the implications of imaginary numbers in relation to time, indicating a potential misunderstanding or miscommunication regarding the problem setup. There is a focus on the conditions under which the quadratic equation yields imaginary roots.

tebes
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Homework Statement


v is the velocity.
v = (5.4t - 4.4t^2)i + 8.8 j
what is the time when v = 10 m/s


Homework Equations





The Attempt at a Solution



I tried the solution the following solution,
( (5.4t - 4.4t^2)^2 + 8.8^2 )^ 1/2 = 10
I solved for t.
But the highest power of t will be up to 4.
Is my method correct in solving the question above?
 
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Square the equation to get rid of the radical. Move the 8.8^2 to RHS leaving only something squared on LHS. Take square root and you are left with quadratic. The problem then is that the roots of the quadratic equation are imaginary. Are you sure you copied the problem correctly. B*B-4*A*C < 0.
 
LawrenceC said:
Square the equation to get rid of the radical. Move the 8.8^2 to RHS leaving only something squared on LHS. Take square root and you are left with quadratic. The problem then is that the roots of the quadratic equation are imaginary. Are you sure you copied the problem correctly. B*B-4*A*C < 0.

yes. I double checked several times already. I was baffling with the same the problem too.
Therefore, i would get more then 1 positive answer, right ?
 
No, you get the square root of a negative number for a time. Last time I looked, time was not an imaginary quantity. B*B-4*A*C is what is under the radical. If it is less than zero, then roots are imaginary.
 
LawrenceC said:
No, you get the square root of a negative number for a time. Last time I looked, time was not an imaginary quantity. B*B-4*A*C is what is under the radical. If it is less than zero, then roots are imaginary.

Is it legit that I move the negative under the radical out ? I would get something like (xxx)^1/2 i ; i = -1. Am i correct ?
 
Sure, you can remove the real part of the number.

(-9)^.5 = 3i where i is (-1)^.5

What bothers me is that the result is an imaginary number and how this relates to the phycality of the problem. If B were larger or if either A or C were different, you would not have a negative under the radical.
 

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