What mistake did I make in my time-dependent force problem solution?

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Loststudent22
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Homework Statement


A time-dependent force,
Fvecbold.gif
= (9.60
ihatbold.gif
− 4.10t
jhatbold.gif
), where
Fvecbold.gif
is in Newtons and t is in seconds, is exerted on a 2.00-kg object initially at rest. (a) At what time will the object be moving with a speed of 15.0 m/s?

Homework Equations


F=ma

The Attempt at a Solution


I solved for a then integrated that and got 4.8t-1.025t^2

Set it equal 15 squared both sides to eliminate the square root when solving for the magnitude and finally got 23.04t^2-1.05t^4 -225=0 and can't solve it. So I must have made some mistake.

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They were the I and J hat vector components. I will have to look into latex I"m not familiar with it yet. I'll trying using a picture next time for the question and my work.
 
Loststudent22 said:
Set it equal 15 squared both sides to eliminate the square root when solving for the magnitude and finally got 23.04t^2-1.05t^4 -225=0 and can't solve it. So I must have made some mistake.
Have you tried a substitution for t? You should be able to turn the equation into a quadratic...
 
I plugged the equation into wolfram to solve for t and it didn't give a answer
 
Loststudent22 said:
I plugged the equation into wolfram to solve for t and it didn't give a answer
You have:

23.04t^2-1.05t^4 -225=0

Let : x = t^2

Write the equation in terms of x. Can you solve that?
 
Yeah it doesn't work it gives an answer with imaginary numbers. I pulled out my 89 and it just says false when I try to use the solve function. Is it possible the online assignment gave me numbers that don't work nicely or I must have made some mistake.
 
Loststudent22 said:
Yeah it doesn't work it gives an answer with imaginary numbers. I pulled out my 89 and it just says false when I try to use the solve function. Is it possible the online assignment gave me numbers that don't work nicely or I must have made some mistake.
So you should revisit how you derived your equation. Lay out the steps so we can take a look.
 
F=ma
a=4.8-2.05t=dv/dt
To arrive at an equation for the instantaneous velocity of object I integrate the above equation

V=4.8t +c1-1.025t^2+c2

The object is at rest when t = 0 s. so c1=c2=0

When V=15 m/s

The magnitude of V=15=√(4.8t)^2+(1.025t^2)^2

I square both sides to eliminate the square root and got the equation I'm having trouble with
 
Loststudent22 said:
F=ma
a=4.8-2.05t=dv/dt
To arrive at an equation for the instantaneous velocity of object I integrate the above equation

V=4.8t +c1-1.025t^2+c2

The object is at rest when t = 0 s. so c1=c2=0

When V=15 m/s

The magnitude of V=15=√(4.8t)^2+(1.025t^2)^2

I square both sides to eliminate the square root and got the equation I'm having trouble with
Really? Because I spot a sign difference right away.
 
Haha woops I see now + instead of -

Thanks