Posted by Sam on Wednesday, November 16, 2011 at 12:30am.
this homework is due in one hour and i have no idea how to answer these 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

Calculus  Steve, Wednesday, November 16, 2011 at 5:50am
speed is reduced by 20ft/sec every second.
So, since 50mph = 73.333 ft/sec, it will take 73.333/20 = 11/3 seconds to reduce the speed to zero
In those 11/3 seconds, the car travels 1/2 at^2 = 1/2 (20)(11/3)^2 = 134.444 ft
_____________________
These kinds of problems just take a bit of thought. If there were no gravity, and no throw, the height would remain constant at
h = 450
If there were no gravity, but the stone were thrown at 8m/s, then the height would be
h = 450  8t
Now, tack on the acceleration due to gravity, and
h = 450  8t  4.9 t^2
Solve for t when h=0
________________________
If f'(x) = 3x^3, then
f(x) = 3/4 x^4 + C
Now the line 81x+y=0 has slope 81
f'(x) = 81 when x = 3
so, the point (3,243) is on the line
so, f(3) = 3/4 * 81 + C = 243
C = 729/4
So, the graph of y = 3/4 x^4 + 729/4 is tangent to the line 81x+y=0 at (3,243)
_______________________
ball 1: h = 432 + 48t  16t^2
ball 2: h = 432 + 24t  16(t1)^2
when is the height the same?
48t  16t^2 = 24t  16t^2 + 32t  16
t = 2
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