Help!!!

[samm_watkins]

this homework is due in one hour and i have no idea how to answer these g.d. questions.

1. A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 20 ft/s^2. What is the distance covered before the car comes to a stop?

2. A stone is dropped from the upper observation deck of a tower, 450 m above the ground. (Assume g = 9.8 m/s^2.)
(A)If the stone is thrown downward with a speed of 8 m/s, how long does it take to reach the ground? (Round your answer to two decimal places.)

3. Find a function f such that f '(x) = 3x^3and the line 81x + y = 0 is tangent to the graph of f

4. Two balls are thrown upward from the edge of a cliff 432 ft above the ground. The first is thrown with a speed of 48 ft/s and the other is thrown a second later with a speed of 24 ft/s.
(A)If they pass each other, give the time when this occurs. If they do not pass each other, enter NONE.

I have seventy percent of the assignment done but i cant figure these out. any help would be awesome

 

[Unknown008]

1. Use the equation $\displaystyle v^2 = u^2 + 2as$

u = 50 mi/hr
a = -20ft/s^2
v = 0 mi/hr

Do the necessary conversions.

2. Use the equation $\displaystyle s = ut + \dfrac12 at^2$

s = 450 m
u = 8 m/s
a = 9.8 m/s^2

3. Integrate f'(x). Then equate the two equations (integral and line given). There might be some way that you'll get a quadratic equation and then you use the fact that for the line to be a tangent, then b^2 - 4ac = 0

4. Get the displacement equation of each of the balls.

$\displaystyle s_1 = ut + \dfrac12 at^2 = 48t - \frac{g}{2}t^2$

$\displaystyle s_2 = ut + \dfrac12 at^2 = 24(t-1) - \frac{g}{2}(t-1)^2$

If their displacement at any time t are the same, then they meet. So, equate the two equations. If there is a solution, that is the time that they meet. If not, they don't meet at all. Don't forget to check the time that you might get with the equation you got. If they meet 'below the cliff' this won't count as meeting.