A 3.50kg box is moving to the right with speed 8.00m/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)=( 6.00N/s2 )t2

Part A

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

Part B

If the force continues to be applied, what is the velocity of the box at 3.50s ?

Express your answer with the appropriate units.

F(t) = (-6.00 N/s^2)t^2

a(t) = (-3.00 N/s^2)t^2  

Because F = ma, the acceleration function is the force function divided by mass (3.50 kg). Because the force is acting to the left, a negative has been introduced.  

Take the integral of the acceleration function with the power rule for integrals. Initial velocity is 8.00 m/s

∫a(t) dt=v(t)+v1

v(t)=(-1m/s^4)*t^3+9 m/s

Setting velocity equal to zero and solving for t.  

v(t)=0

t^3=9s^3

t=∛9s

=2.08 s

The integral of velocity is position. The object begins at the origin so initial position is 0  

∫v(t) dt=  x(t)

x(t)=(-0.25m/s^4)*t^4+(9m/s)*t

Plugging the t from step 3 into the x(t) function from step 4. This is the answer to part a.  

x(2.08)=14 m

plug 3.50 s into the velocity function from step 2. Speed is the absolute value of velocity. This is the answer to part b.  

v(3.5)=(1 m/s^4)(3.5 s)^3+9 m/s

        = -18 m/s

speed(3.5 s)=║v(3.5)║=18 m/s

i think it's 35 meters