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

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
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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
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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)
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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

I'd be happy to help you with your homework questions! Let's break down each question and explain how to find the answers.

1. To find the distance covered before the car comes to a stop, you need to use the equations of motion. The equation you need here is the one that relates distance, initial velocity, final velocity, and acceleration:

v^2 = u^2 + 2as

Where:
- v is the final velocity (which is 0 when the car comes to a stop),
- u is the initial velocity (50 mi/h),
- a is the acceleration due to braking (-20 ft/s^2), and
- s is the distance traveled.

Rearranging the equation to solve for distance (s), we have:

s = (v^2 - u^2) / (2a)

Plugging in the values: v = 0, u = 50 mi/h, and a = -20 ft/s^2, remember to convert the units to a consistent one (either miles or feet) to obtain the correct answer.

2. In this question, you need to find how long it takes for the stone to reach the ground when it is dropped from a height of 450 m or thrown downward with a speed of 8 m/s. Since the stone is only affected by the force of gravity, we can use the equations of motion for free-fall:

h = ut + (1/2)gt^2

Where:
- h is the height (450 m),
- u is the initial velocity (8 m/s),
- g is the acceleration due to gravity (9.8 m/s^2),
- t is the time it takes to reach the ground.

Rearranging the equation to solve for time (t), we have a quadratic equation:

(1/2)gt^2 + ut - h = 0

You can solve this quadratic equation for time using the quadratic formula or factoring.

3. Given the derivative f'(x) = 3x^3 and the fact that the line 81x + y = 0 is tangent to the graph of f, we can find the function f. To find f(x), we need to integrate f'(x):

∫f'(x) dx = ∫3x^3 dx

Integrating 3x^3, we get:

f(x) = (3/4) x^4 + C

To find the constant C, we need to use the condition that the line 81x + y = 0 is tangent to the graph of f. This means that the slope of the tangent line should be equal to the slope of the function f(x) at the point of tangency. The slope of the line is -81, so we equate it to the derivative of f(x) and solve for x:

f'(x) = -81

3x^3 = -81

Solve for x: x = -3

Since the line is tangent to the graph of f, we can substitute x = -3 into f(x) to find C:

f(-3) = (3/4)(-3)^4 + C

Solve for C.

4. In this question, you need to determine whether the two balls pass each other and, if so, find the time when this occurs. To solve this problem, you need to calculate the time it takes for each ball to reach the top of its trajectory by using the equation of motion:

v = u + at

Where:
- v is the final velocity,
- u is the initial velocity,
- a is the acceleration due to gravity (-32 ft/s^2, considering upward motion as positive),
- t is the time taken.

First, calculate the time it takes for the first ball to reach its maximum height by plugging in the values: v = 0 (at maximum height), u = 48 ft/s, a = -32 ft/s^2.

Repeat the same process to find the time it takes for the second ball to reach its maximum height, but this time u = 24 ft/s and a = -32 ft/s^2.

If the time for the first ball to reach its maximum height is greater than the time for the second ball to reach its maximum height, then the balls pass each other at some point. You can subtract the time for the second ball from the time for the first ball to find the exact time when they pass each other. Otherwise, if the time for the second ball to reach its maximum height is equal to or greater than the time for the first ball, then they do not pass each other (enter NONE).

Remember to carefully perform the necessary unit conversions throughout these calculations to ensure you obtain the correct answers. Good luck with your homework!