Question

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.

Answer 1

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

s(t) = -2sin(t) + ½t² + 2t + 3

Average value of s: 17.0 (3 sf)

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

Answer 2

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

position s(t) =
`-2sin(t)+(t^2)/(2) +2t + 3`

average of position from 2 to 5 = 17.666

Explanation:

velocity is anti-derivative/integral of acceleration

`\int{(2sin(t)+1)} \, dt` = -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

`\int{(-2cos(t)+t+2)} \, dt` =
`-2sin(t)+(t^2)/(2) +2t + C`

s(0) = 3

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

3 = C

s(t) =
`-2sin(t)+(t^2)/(2) +2t + 3`

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

`(1)/(b-a) \int\limits^b_a {f(x)} \, dx`

average of position from t = 2 to t = 5

`(1)/(5-2) \int\limits^5_2 {(-2sin(t)+(t^2)/(2) +2t + 3)} \, dt`

`\int\limits^5_2 {(-2sin(t)+(t^2)/(2) +2t + 3)} \, dt` = 2cos(5)+125/6+25+15 - 2cos(2) - 8/6 -4 + 6

=52.999 (use a calculator)

52.999/3 = 17.666