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

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Homework Help Overview

The discussion revolves around a time-dependent force problem involving a 2.00-kg object initially at rest, with a force expressed as F = (9.60 - 4.10t) Newtons. The original poster seeks to determine the time at which the object reaches a speed of 15.0 m/s.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the integration of acceleration to find velocity and the subsequent squaring of terms to eliminate square roots. There are attempts to reformulate the resulting equation into a solvable form, including suggestions for substitutions and questioning the validity of the derived equation.

Discussion Status

Some participants have offered guidance on reformulating the equation and checking for errors in the derivation process. There is an acknowledgment of potential issues with the numbers provided in the assignment, and participants are exploring various interpretations of the problem setup.

Contextual Notes

Participants note that the original poster may have made a mistake in their calculations or assumptions, and there is a mention of the possibility that the online assignment may have provided challenging numbers. The discussion includes references to using tools like Wolfram for solving equations, which did not yield satisfactory results.

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.

[ URLs fixed by a mentor ]
 
Last edited by a moderator:
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Loststudent22, you didn't have the correct URLs for the i and j characters at your webassign site. I fixed them.

Next time you can try latex. ##{\hat{i}}\ {\hat{j}}##
 
Last edited:
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
 
  • #10
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.
 
  • #11
Haha woops I see now + instead of -

Thanks
 

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