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(t-1)^2

when is the height the same?

48t - 16t^2 = 24t - 16t^2 + 32t - 16

t = 2

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