1) please show another example of 1 so I can try it out myself. Provide a answer key. 2) explain in detail how to solve it. Another example would be helpful. 3) please explain in detail plus show the formula. Provide another example. 4) explain in detail and what's the difference of 4 and 5? Give another example. 5) explain why and what formulas to use.

A Alternative formats A Alternative formats Page of9 0 - 200M + of 11 0 - 200M Motion in a gravitational Field Population Growth Rates the position of the stone ( measured as the height above themvery after thecond; and it's realise where no us the fever at the ter A. Find the velocity and acceleration functions. The population of a culture of cells increases and approach a Step 1. Velocity is the derivative of position. constant level. The population is modeled by the function p(t) = 400 -2, where t 2 0 is measured in hours. (-1612 + 641 + 96) = -32t + 64 Step 2. Acceleration is the derivative of position. Compute the instantaneous growth rate of the population for t 20. " = (-32t + 64) = -32 Step 1. Take the derivative of our function using the quotient B. What is the highest point above the ver reached by the stone. rule. (write out the terms for quotient rule) Step 3. The highest point will be when the velocity reaches O (goes higher to lower) -32t + 64 = 0 = 321 = 64 m t = 2. f (t) = 12 + 1g(t) = +2 +4 f'(t) = 2t g'(t) = 2t Step 4. Plug 2 into the position equation to find the maximum height. Step 2. Take the derivative "? = 400 ( +4)2t-(12+1)2t -16(2)3 + 64(2) +96=-64 +128 +96 = 160 ft. (t2+4)2 Step 3. Simplify. 400 2 +8t-213-2t 2400t (12+4)2 (12+4)2 Q Search 490 - 200M + Velocity and Acceleration Suppose the position (in feet) of an object mov izontally at time t (in seconds) is set-5t for Destees Motion in a gravitational Field A. Graph the velocity function on the interval 0 S t S 5, and determine when the object is stationary, moving to the left and moving to the right. Suppose a stone is thrown vertically upward with an initial velocity of 64ft/s from a To find the velocity we take the derivative of the movement , (t2 - 5t) = 2t - 5 and graph 2 bridge 96ft above a river. By Newton's laws of motion, the position of the stone ver. By Newton's laws of motion (measured as the height above the river) after t seconds is s(t)=-16 t2+64t+96 where s=0 It is to the left from [0,5/2) stationary at 5/2 and moving to the right (5/2,5] is the level of the river. B. Graph the acceleration function from 0 S t s 5 celeration when v=0. . With what velocity will the rock hit the water. To find the acceleration we take the second deriva e derivative of the velocity. 417 = 2 Step 5. Find the time that the rock will hit the river by solving the height equation for 0. -16t2 + 64t + 96 =0 = -16(t2 - 4t - 6) with the quadratic formula we get t=- 1.16 and 5.16. We cannot use negative time so t=5.16. Step 6. Plug 5.16 into the velocity equation v(5.16)=-32(5.16)+64=-101.12 ft per second Q Search LOC-ROBO 249667 /files?preview=34742593 Pape 69 0 - 200M + Elasticity in pork prices The demand for processed pork in Canada is described by the function Dip)-286-20p. A. Compute the price elasticity of the demand Step 1. Plug the demand equation in for demand in the elasticity equation. (286 - 20p) (206-205) Step 2. Take the derivative and plug in into equation. 3 20P (286 - 20p) - -20. -286 - 20p Step 3. Simplify. 10p elastic and mefasten