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?
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:
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
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!