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