Physics (velocity, acceleration, displacement, gravity,...) Please SHOW WORK/STEPS/EXPLANATIONS?

Question

In an arcade video game, a spot is programmed to move across the screen according to x = 8.00t - 0.800t3, where x is distance in centimeters measured from the left edge of the screen and t is time in seconds. When the spot reaches a screen edge, at either x = 0 or x = 15.0 cm, t is reset to 0 and the spot starts moving again according to x(t).
(f) When does it first reach an edge of the screen after t = 0?

A motorcyclist who is moving along an x axis directed toward the east has an acceleration given by a = (6.1 - 2.0t) m/s2 for 0 t 6.0 s. At t = 0, the velocity and position of the cyclist are 3.2 m/s and 7.3 m.
(a) What is the maximum speed achieved by the cyclist? (m/s)
(b) What total distance does the cyclist travel between t = 0 and 6.0 s? (m)

To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 1.60 m. Suppose the ball is in contact with the floor for 13.5 ms.
(a) What is the magnitude of its average acceleration during the contact with the floor?

An object falls from height h from rest and travels 0.49h in the last 1.00 s.
(a) Find the time of its fall.
(b) Find the height of its fall.

A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot was in view for a total of 0.39 s, and the top-to-bottom height of the window is 1.65 m. How high above the window top did the flowerpot go? (m)

A lead ball is dropped into a lake from a diving board 5.10 m above the water. It hits the water with a certain velocity and then sinks to the bottom with this same constant velocity. It reaches the bottom 4.84 s after it is dropped. (Assume the positive direction is upward.)
(a) How deep is the lake?

I HAVE NO IDEA HOW TO DO THESE! :(
I SPENT A TOTAL OF 10 HOURS ON THESE! ARGH

NO INTEGRATION NEEDED! (ONLY NEEDS BASIC DERIVATIVES)

DO ANY ONE (DO MORE IF U FEEL LIKE IT... not telling u that u should do all)

Answer 1

Arcade:
Since the spot can travel 15 cm to the other side, we can set
15 = 8*tf -0.8tf^3, where tf is the time to travel.
We can move all terms to the left to get a cubic equation. One root should be positive, which would be tf.

maximum velocity => dv/dt=0. Since dv/dt=a, we set a=0 and 2.0 tm = 6.1. Solve for tm
Then v = vo + a* tm. Substitute in for vo, a and tm.
You will have a quadratic to solve.
Finally, x=xo+vo*t+(a/2)*t^2. You will have a cubic eqtn to solve (no rest for the wicked sayeth God).

That's enough problems for one entry. If you pile it on, no one is going to spend the time to explain or solve for 10 points.

Answer 2

The initial velocity is bigger than the wanted final velocity. (3E5 m/s) > (5.4E4 m/s) As the acceleration is positive (8E14 m/s²), you will never get that final velocity. If the acceleration were negative (-8E14 m/s²), then 5.4E4 = 3E5 + t ∙ (-8E14) t = 3E-10 s (t = 0.3 ns (nanoseconds)) The distance will be d = (5.4E4 + 3E5) ∙ (3E-10) / 2 d = 0.54E-4 m (d = 54 nm (nanometer))