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The acceleration of an object is =4 m/s^2.?
The acceleration of an object is =4 m/s^2. Find the distance of the object from the origin under the initial conditions (0)=9 m, (0)=16 m/s.

(Use symbolic notation and fractions where needed.)

The acceleration of an object is 4 ms2 The acceleration of an object is 4 ms2 Find the distance of the object from the origin under the initial conditions 09 m class=

Respuesta :

Space

Answer:

[tex]\displaystyle s(t) = \frac{2t^3}{3} + 16t + 9[/tex]

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Calculus

Antiderivatives - Integrals

Integration Constant C

Integration Rule [Reverse Power Rule]:                                                               [tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]

Integration Property [Multiplied Constant]:                                                         [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Integration Property [Addition/Subtraction]:                                                       [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]

Step-by-step explanation:

*Note:

Velocity is the derivative of Position, and Acceleration is derivative of Velocity.

↓

Velocity is integration of Acceleration, Position is integration of Velocity.

Step 1: Define

a(t) = 4t m/s²

s(0) = 9 m

v(0) = 16 m/s

Step 2: Find Velocity Function

Integration Pt. 1

  1. [Velocity] Set up integral:                                                                             [tex]\displaystyle v(t) = \int {a(t)} \, dt[/tex]
  2. [Velocity] Substitute in function:                                                                   [tex]\displaystyle v(t) = \int {4t} \, dt[/tex]
  3. [Velocity] Rewrite [Integration Property - Multiplied Constant]:                 [tex]\displaystyle v(t) = 4\int {t} \, dt[/tex]
  4. [Velocity] Integrate [Integration Rule - Reverse Power Rule]:                   [tex]\displaystyle v(t) = 4(\frac{t^2}{2}) + C[/tex]
  5. [Velocity] Multiply:                                                                                         [tex]\displaystyle v(t) = 2t^2 + C[/tex]

Finding C

  1. [Velocity] Substitute in initial condition:                                                       [tex]\displaystyle v(0) = 2(0)^2 + C[/tex]
  2. [Velocity] Substitute in function value:                                                         [tex]\displaystyle 16 = 2(0)^2 + C[/tex]
  3. [Velocity] Evaluate exponents:                                                                     [tex]\displaystyle 16 = 2(0) + C[/tex]
  4. [Velocity] Multiply:                                                                                         [tex]\displaystyle 16 = C[/tex]
  5. Rewrite:                                                                                                         [tex]\displaystyle C = 16[/tex]

Velocity Function: [tex]\displaystyle v(t) = 2t^2 + 16[/tex]

Step 3: Find Position Function

Integration Pt. 2

  1. [Position] Set up integral:                                                                             [tex]\displaystyle s(t) = \int {v(t)} \, dt[/tex]
  2. [Position] Substitute in function:                                                                   [tex]\displaystyle s(t) = \int {2t^2 + 16} \, dt[/tex]
  3. [Position] Rewrite [Integration Property - Addition]:                                    [tex]\displaystyle s(t) = \int {2t^2} \, dt + \int {16} \, dt[/tex]
  4. [Position] Rewrite [Integration Property - Multiplied Constant]:                 [tex]\displaystyle s(t) = 2\int {t^2} \, dt + 16\int {} \, dt[/tex]
  5. [Position] Integrate [Integration Rule - Reverse Power Rule]:                   [tex]\displaystyle s(t) = 2(\frac{t^3}{3}) + 16t + C[/tex]
  6. [Position] Multiply:                                                                                         [tex]\displaystyle s(t) = \frac{2t^3}{3} + 16t + C[/tex]

Finding C

  1. [Position] Substitute in initial condition:                                                       [tex]\displaystyle s(0) = \frac{2(0)^3}{3} + 16(0) + C[/tex]
  2. [Position] Substitute in function value:                                                         [tex]\displaystyle 9 = \frac{2(0)^3}{3} + 16(0) + C[/tex]
  3. [Position] Evaluate exponents:                                                                     [tex]\displaystyle 9 = \frac{2(0)}{3} + 16(0) + C[/tex]
  4. [Position] Multiply:                                                                                         [tex]\displaystyle 9 = \frac{0}{3} + C[/tex]
  5. [Position] Divide:                                                                                           [tex]\displaystyle 9 = C[/tex]
  6. [Position] Rewrite:                                                                                         [tex]\displaystyle C = 9[/tex]

Position Function: [tex]\displaystyle s(t) = \frac{2t^3}{3} + 16t + 9[/tex]

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Integration

Book: College Calculus 10e

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