Respuesta :
Answer:
[tex]\displaystyle s(t) = 5t^2 - 20t + 8[/tex]
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
Algebra I
Functions
- Function Notation
Calculus
Differentiation
- Derivatives
- Derivative Notation
- Position/Velocity/Acceleration
Integration
- Integrals
- Indefinite Integrals
- Integration Constant C
- Position/Velocity/Acceleration
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:
Step 1: Define
Identify
a(t) = 10 ft/sec²
v(0) = -20 ft/sec
s(0) = 8 ft
Step 2: Find Velocity Function
- Define: [tex]\displaystyle v(t) = \int {a(t)} \, dt[/tex]
- [Integrand] Substitute in acceleration function: [tex]\displaystyle v(t) = \int {10} \, dt[/tex]
- [Integral] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle v(t) = 10\int {} \, dt[/tex]
- [Integral] Integrate [Integration Rule - Reverse Power Rule]: [tex]\displaystyle v(t) = 10t + C[/tex]
- Substitute in t [Velocity Function v(t)]: [tex]\displaystyle v(0) = 10(0) + C[/tex]
- Substitute in function values: [tex]\displaystyle -20 = 10(0) + C[/tex]
- Solve: [tex]\displaystyle C = -20[/tex]
- Substitute in C [Velocity Function v(t)]: [tex]\displaystyle v(t) = 10t - 20[/tex]
Step 3: Find Position Function
- Define: [tex]\displaystyle s(t) = \int {v(t)} \, dt[/tex]
- [Integrand] Substitute in velocity function: [tex]\displaystyle s(t) = \int {(10t - 20)} \, dt[/tex]
- [Integral] Rewrite [Integration Property - Addition/Subtraction]: [tex]\displaystyle s(t) = \int {10t} \, dt - \int {20} \, dt[/tex]
- [Integrals] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle s(t) = 10\int {t} \, dt - 20\int {} \, dt[/tex]
- [Integrals] Integrate [Integration Rule - Reverse Power Rule]: [tex]\displaystyle s(t) = 10(\frac{t^2}{2}) - 20t + C[/tex]
- Simplify: [tex]\displaystyle s(t) = 5t^2 - 20t + C[/tex]
- Substitute in t [Position Function s(t)]: [tex]\displaystyle s(0) = 5(0)^2 - 20(0) + C[/tex]
- Substitute in function values: [tex]\displaystyle 8 = 5(0)^2 - 20(0) + C[/tex]
- Solve: [tex]\displaystyle C = 8[/tex]
- Substitute in C [Position Function s(t)]: [tex]\displaystyle s(t) = 5t^2 - 20t + 8[/tex]
∴ our position function is [tex]\displaystyle s(t) = 5t^2 - 20t + 8[/tex].
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Book: College Calculus 10e