# Time-dependent Force problem

1. Feb 20, 2015

### Loststudent22

1. The problem statement, all variables and given/known data
A time-dependent force, = (9.60 − 4.10t ), where 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?

2. Relevant equations
F=ma

3. 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: Apr 19, 2017
2. Feb 21, 2015

### Staff: Mentor

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: Feb 21, 2015
3. Feb 21, 2015

### Loststudent22

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.

4. Feb 21, 2015

### Staff: Mentor

Have you tried a substitution for t? You should be able to turn the equation into a quadratic...

5. Feb 21, 2015

### Loststudent22

I plugged the equation into wolfram to solve for t and it didn't give a answer

6. Feb 21, 2015

### Staff: Mentor

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?

7. Feb 21, 2015

### Loststudent22

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.

8. Feb 21, 2015

### Staff: Mentor

So you should revisit how you derived your equation. Lay out the steps so we can take a look.

9. Feb 21, 2015

### Loststudent22

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. Feb 21, 2015

### Staff: Mentor

Really? Because I spot a sign difference right away.

11. Feb 21, 2015

### Loststudent22

Haha woops I see now + instead of -

Thanks