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## Homework Statement

A 2.70 kg box is moving to the right with speed 9.50 m/s on a horizontal, frictionless surface. At t = 0 a horizontal force is applied to the box. The force is directed to the left and has magnitude F(t)=( 6.00 N/s2 )t2

What distance does the box move from its position at t=0 before its speed is reduced to zero?

## Homework Equations

F = ma

vf^2 = vi^2 + 2ad

xf = xi + vit + .5axt^2

## The Attempt at a Solution

So i am confused about this F(t) = 6.00(n/s2)t^2

I've read solutions to problems similar to this where they say to integrate to get a velocity equation and then solve for v=0.

So I did that:

a=6t^2

v=(6t^3)/3 + C, and plugged in my initial velocity, 9.50 m/s as C. I am not sure why but I saw someone do this in the solutions I looked at.

From this I solved for time and got 1.68s.

Then I used Newton's 2nd Law .

F = ma = 6.00t^2. I didn't include weight and normal force because the object is only moving along the x axis.

ma = 6.00t^2

2.70a=6.00(1.68^2)

a = 6.272m/s^2

Now that I have acceleration, initial velocity, initial position, and time, I use my kinematic equation to find the final position where Vf is 0.

xf = 0 + 9.5(1.68) + .5(6.272)(1.68^2)

xf= 15.96 + 8.85

xf = 24.8m

This is wrong. The parts where I am very confused are interpreting what F(t) = 6.00(N/s2)t^2 means compared to F(t) = 6.00N. I am also confused with knowing how to determine what C is, but maybe I would understand that part if I understood the force function. Thank you for looking at this!