# Find the time when velocity = 10 m/s

1. Oct 14, 2011

### tebes

1. The problem statement, all variables and given/known data
v is the velocity.
v = (5.4t - 4.4t^2)i + 8.8 j
what is the time when v = 10 m/s

2. Relevant equations

3. 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?

2. Oct 14, 2011

### LawrenceC

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.

3. Oct 14, 2011

### tebes

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 ?

4. Oct 14, 2011

### LawrenceC

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.

5. Oct 14, 2011

### tebes

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 ?

6. Oct 15, 2011

### LawrenceC

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.