Answer:
t = 1.068 s
Step-by-step explanation:
given,
a(t) =- k v(t)
speed of the object decreases from 800 ft/s to 700 ft/s
distance = 1400 ft
time for deceleration = ?
a(t) =- k v(t)
![\dfrac{dv}{dt}= - kv](https://tex.z-dn.net/?f=%5Cdfrac%7Bdv%7D%7Bdt%7D%3D%20-%20kv)
![\dfrac{dv}{v}= - kdt](https://tex.z-dn.net/?f=%5Cdfrac%7Bdv%7D%7Bv%7D%3D%20-%20kdt)
integrating both side
![\int_{800}^{700}\dfrac{dv}{v}= - k\int dt](https://tex.z-dn.net/?f=%5Cint_%7B800%7D%5E%7B700%7D%5Cdfrac%7Bdv%7D%7Bv%7D%3D%20-%20k%5Cint%20dt)
![ln(\dfrac{7}{8})=-kt](https://tex.z-dn.net/?f=ln%28%5Cdfrac%7B7%7D%7B8%7D%29%3D-kt)
..........(1)
since,
![v = v_1e^{-kt}](https://tex.z-dn.net/?f=v%20%3D%20v_1e%5E%7B-kt%7D)
![\dfrac{ds}{dt} = 800e^{-kt}](https://tex.z-dn.net/?f=%5Cdfrac%7Bds%7D%7Bdt%7D%20%3D%20800e%5E%7B-kt%7D)
![\int ds= 800\int_0^t e^{-kt}dt](https://tex.z-dn.net/?f=%5Cint%20ds%3D%20800%5Cint_0%5Et%20e%5E%7B-kt%7Ddt)
![1400= -\dfrac{800}{k}[e^{-kt}-e^0]](https://tex.z-dn.net/?f=1400%3D%20-%5Cdfrac%7B800%7D%7Bk%7D%5Be%5E%7B-kt%7D-e%5E0%5D)
from equation 1
![k= -\dfrac{8}{14}[\dfrac{7}{8}-1]](https://tex.z-dn.net/?f=k%3D%20-%5Cdfrac%7B8%7D%7B14%7D%5B%5Cdfrac%7B7%7D%7B8%7D-1%5D)
![k= \dfrac{1}{8}](https://tex.z-dn.net/?f=k%3D%20%5Cdfrac%7B1%7D%7B8%7D)
putting value of k in equation (1)
![t = -8ln(\dfrac{7}{8})](https://tex.z-dn.net/?f=t%20%3D%20-8ln%28%5Cdfrac%7B7%7D%7B8%7D%29)
t = 1.068 s
Let the width path be x.
Length of the outer rectangle = 26 + 2x.
Width of the outer rectangle = 8 +2x.
Combined Area = (2x + 26)*(2x + 8) = 1008
2x*(2x + 8) + 26*(2x + 8 ) = 1008
4x² + 16x + 52x + 208 = 1008
4x² + 68x + 208 - 1008 = 0
4x² + 68x - 800 = 0. Divide through by 4.
x² + 17x - 200 = 0 . This is a quadratic equation.
Multiply first and last coefficients: 1*-200 = -200
We look for two numbers that multiply to give -200, and add to give +17
Those two numbers are 25 and -8.
Check: 25*-8 = -200 25 + -8 = 17
We replace the middle term of +17x in the quadratic expression with 25x -8x
x² +17x - 200 = 0
x² + 25x - 8x - 200 = 0
x(x + 25) - 8(x + 25) = 0
(x+25)(x -8) = 0
x + 25 = 0 or x - 8 = 0
x = 0 -25 x = 0 + 8
x = -25 x = 8
The width of the path can not be negative.
The only valid solution is x = 8.
The width of the path is 8 meters.
Step-by-step explanation:
![- 2 < 5 - \frac{k}{2} \\ - 7 < - \frac{k}{2} \\ \frac{k}{2} > 7 \\ k > 14](https://tex.z-dn.net/?f=%20-%202%20%3C%205%20-%20%20%5Cfrac%7Bk%7D%7B2%7D%20%20%5C%5C%20%20-%207%20%3C%20%20-%20%20%5Cfrac%7Bk%7D%7B2%7D%20%20%5C%5C%20%20%5Cfrac%7Bk%7D%7B2%7D%20%20%3E%207%20%5C%5C%20k%20%3E%2014)