Problem listed below.......its a calculus 1 problem/physics!?

Question

A car traveling at 48ft/sec decelerates at a constant 4feet per second squared.

Let denote f(x)the position function seconds after the decelerating process started. Then . f(0)=0
What is f'(0)? Answer:

What is f''(x)? Answer:

Use the above information to find f'(x)

Then, use f(0)=0 to find f(x)

How many feet does the car travel before coming to a complete stop?

Answer:

Answer 1

What is f'(0)? Answer: The initial velocity of the car.

What is f''(x)? Answer: The rate of deceleration. (don't forget the -)


Use the above information to find f'(x)
See if you can. It doesn't help you for us to do it for you.

Think a little.

Edit...I'll help you since the next guy tried to do everything.

f ' (x) isn't 4t it is 48 - 4t, which is the initial velocity plus the deceleration

This would make f(x) = 48t - 2t^2

How long does it take to stop?

Set f ' (x) = 0

48 - 4t = 0
4t = 48
t = 12

Now plug 12 into f(x)

f(x) = 48t - 2t^2
f(12) = 48(12) - 2(12)^2
f(12) = 576 - 288
f(12) = 288

Answer 2

What is f'(0)? Answer: - 4 ft/s^2

What is f''(x)? Answer: 48 ft/s

Use the above information to find f'(x)
f'(x) = 48 - 4x

Then, use f(0)=0 to find f(x)
f(x) = 48x - 2x^2

How many feet does the car travel before coming to a complete stop?
0 = 48 - 4x
4x = 48
x = 12 s
f(x) = 48(12) - 2(12)^2
f(x) = 288 ft.

Answer 3

f '(0) = v_0 = initial velocity = 48 [ ft/sec ]

f ''(x) = a = acceleration = -4 [ ft/sec^2 ]

f '(x) = int f ''(x) dx = -4x + c

f '(0) = c and f'(0) = 48, so c = 48

f(x) = int f '(x) dx = int (-4x + 48) dx = -2x^2 + 48x + c (new c)

f(0) = c and f(0) = 0, so c = 0 and f(x) = -2x^2 + 48x

For the last part you need to solve v(x) = f '(x) = 0

Answer 4

a = -4
v = 48

f(0) = 0
f'(0) = 0
f"(x) = -4

f'(x) = -4t
f(x) = -2t^2

Vo = 48
a = -4
V = 0
x = ??

V^2 = Vo^2 + 2ax
0 = (48)^2 + 2(-4)x
-2304 = -8x
x = 288 ft