The acceleration, in feet per second per second, of an object is given by the acceleration function a(t)=2 sin t+1. The initial velocity is v(0)=0 and the initial position is s(0) =3. Find the equation of velocity function. Find the position function and the average value of the position function from time t = 2 seconds to t = 5 seconds. Show all your work.

velocity v(t) = -2cos(t)+t+2

position s(t) =

average of position from 2 to 5 = 17.666

Step-by-step explanation:

velocity is anti-derivative/integral of acceleration

= -2cos(t)+t+C

v(0) = 0

0 = -2cos(0)+0+C = -2 +C; C = 2

v(t) = -2cos(t)+t+2

position is anti-derivative of velocity

=

s(0) = 3

3 = -2sin(0) + 0 + 0+ C

3 = C

s(t) =

average value of a function of a domain [a, b] is given by the equation

average of position from t = 2 to t = 5

= 2cos(5)+125/6+25+15 - 2cos(2) - 8/6 -4 + 6

=52.999 (use a calculator)

52.999/3 = 17.666

v(t) = v(t) = -2cos(t) + t + 2

s(t) = -2sin(t) + ½t² + 2t + 3

Average value of s: 17.0 (3 sf)

Step-by-step explanation:

v(t) is the integral of a(t)

v(t) = -2cos(t) + t + c

t = 0, v = 0

0 = -2 + 0 + c

c = 2

v(t) = -2cos(t) + t + 2

Displacement/position is the integral of v(t)

s(t) = -2sin(t) + ½t² + 2t + c

t = 0, s = 3

3 = 0 + 0 + 0+ c

c = 3

s(t) = -2sin(t) + ½t² + 2t + 3

Integral of s:

2cos(t) + ⅙t³ + t² + 3t + c

Average value

= 1/(b - a) × integral

= 1/(5-2) × [ (2cos(5) + ⅙(5)³ + 5² + 3(5) + c) - (2cos(2) + ⅙(2)³ + 2² + 3(2) + c) ]

= ⅓[61.4006577 - 10.50103966]

= 16.96653935

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

where is the graph, sweetheart?

step-by-step explanation: