AP Calculus BC 8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals Study Notes - New Syllabus
AP Calculus BC 8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals Study Notes- New syllabus
AP Calculus BC 8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Definite integrals allow us to solve problems involving the accumulation of change over an interval.
Key Concepts:
- Connecting Position, Velocity, and Acceleration of Functions Using Integrals
Connecting Position, Velocity, and Acceleration of Functions Using Integrals
Connecting Position, Velocity, and Acceleration of Functions Using Integrals
1. Connecting Position and Velocity Using Integrals
If an object moves along a straight line with velocity \( v(t) \) at time \( t \), then its position function \( s(t) \) can be found using the relationship between velocity and position:
\( s(t) = s(t_0) + \displaystyle \int_{t_0}^{t} v(x) \, dx \)
Here:
- \( s(t) \) = position at time \( t \)
- \( s(t_0) \) = initial position at time \( t_0 \)
- \( v(x) \) = velocity function
- The integral represents the net change in position between \( t_0 \) and \( t \).
Key Idea: The velocity function is the derivative of the position function, and integrating velocity over a time interval gives the net change in position.
Example :
An object moves along a line with velocity \( v(t) = 3t^2 – 2t \) meters per second. If its position at \( t = 1 \) second is \( s(1) = 5 \) meters, find \( s(4) \).
▶️ Answer/Explanation
\( s(4) = s(1) + \displaystyle \int_{1}^{4} (3t^2 – 2t) \, dt \)
\( = 5 + \left[ t^3 – t^2 \right]_{1}^{4} \)
\( = 5 + \left[ (64 – 16) – (1 – 1) \right] \)
\( = 5 + (48 – 0) = 53 \ \text{meters} \)
Final Answer: The object’s position at \( t = 4 \) is \( 53 \) meters.
2. Connecting Velocity and Acceleration Using Integrals
If an object’s acceleration function \( a(t) \) is known, its velocity function \( v(t) \) can be found using the relationship between acceleration and velocity:
\( v(t) = v(t_0) + \displaystyle \int_{t_0}^{t} a(x) \, dx \)
Here:
- \( v(t) \) = velocity at time \( t \)
- \( v(t_0) \) = initial velocity at time \( t_0 \)
- \( a(x) \) = acceleration function
- The integral represents the net change in velocity between \( t_0 \) and \( t \).
Key Idea: The acceleration function is the derivative of the velocity function, and integrating acceleration over a time interval gives the net change in velocity.
Example:
A particle moves along a line with acceleration \( a(t) = 4t – 2 \) meters per second squared. If \( v(0) = 3 \) m/s, find \( v(5) \).
▶️ Answer/Explanation
\( v(5) = v(0) + \displaystyle \int_{0}^{5} (4t – 2) \, dt \)
\( = 3 + \left[ 2t^2 – 2t \right]_{0}^{5} \)
\( = 3 + \left[ (50 – 10) – (0 – 0) \right] \)
\( = 3 + (40 – 0) = 43 \ \text{m/s} \)
Final Answer: The particle’s velocity at \( t = 5 \) is \( 43 \) m/s.
3. Connecting Position and Acceleration Using Two Integrals
If an object’s acceleration function \( a(t) \) is known, its position function \( s(t) \) can be found by integrating twice, because acceleration is the second derivative of position:
\( v(t) = v(t_0) + \displaystyle \int_{t_0}^{t} a(x) \, dx \)
\( s(t) = s(t_0) + \displaystyle \int_{t_0}^{t} v(x) \, dx \)
Substituting the first equation into the second allows you to write position directly in terms of acceleration:
\( s(t) = s(t_0) + \displaystyle \int_{t_0}^{t} \left[ v(t_0) + \displaystyle \int_{t_0}^{x} a(u) \, du \right] dx \)
Key Idea: The first integration of \( a(t) \) gives the velocity function (up to an initial velocity constant), and the second integration gives the position function (up to an initial position constant).
Example:
A particle has acceleration \( a(t) = 6t \) m/s², initial velocity \( v(0) = 4 \) m/s, and initial position \( s(0) = 10 \) m. Find its position at \( t = 3 \).
▶️ Answer/Explanation
\( v(t) = v(0) + \displaystyle \int_{0}^{t} 6x \, dx \)
\( = 4 + \left[ 3x^2 \right]_{0}^{t} \)
\( = 4 + 3t^2 \)
\( s(t) = s(0) + \displaystyle \int_{0}^{t} \left[ 4 + 3x^2 \right] dx \)
\( = 10 + \left[ 4x + x^3 \right]_{0}^{t} \)
\( = 10 + (4t + t^3) \)
\( s(3) = 10 + (4(3) + 27) = 10 + (12 + 27) = 49 \ \text{m} \)
Final Answer: The particle’s position at \( t = 3 \) is \( 49 \) meters.
Example:
A particle moves along a straight line so that its velocity is given by \( v(t) = t^2 – 4t + 3 \) m/s, where \( t \) is measured in seconds. The particle’s initial position at \( t = 0 \) is \( s(0) = 2 \) meters.
(a) Find the particle’s position function \( s(t) \).
(b) Find the acceleration function \( a(t) \).
(c) Determine the total distance traveled by the particle on the interval \( 0 \le t \le 4 \).
▶️ Answer/Explanation
(a)
\( s(t) = s(0) + \displaystyle \int_{0}^{t} v(x) \, dx \)
\( = 2 + \displaystyle \int_{0}^{t} \left( x^2 – 4x + 3 \right) dx \)
\( = 2 + \left[ \dfrac{x^3}{3} – 2x^2 + 3x \right]_{0}^{t} \)
\( = 2 + \dfrac{t^3}{3} – 2t^2 + 3t \)
(b)
\( a(t) = v'(t) = 2t – 4 \)
(c)
We must find where velocity changes sign on \( [0, 4] \) to break into intervals.
\( v(t) = t^2 – 4t + 3 = (t – 1)(t – 3) \)
Zeros at \( t = 1 \) and \( t = 3 \).
Distance = sum of absolute displacements over each interval.
Interval 1: \( [0, 1] \)
\( s(1) = 2 + \dfrac{1^3}{3} – 2(1)^2 + 3(1) = 2 + \dfrac{1}{3} – 2 + 3 = \dfrac{10}{3} \)
Displacement: \( \dfrac{10}{3} – 2 = \dfrac{4}{3} \) (positive)
Interval 2: \( [1, 3] \)
\( s(3) = 2 + \dfrac{27}{3} – 18 + 9 = 2 + 9 – 18 + 9 = 2 \)
Displacement: \( 2 – \dfrac{10}{3} = -\dfrac{4}{3} \) (negative, so distance is \( \dfrac{4}{3} \))
Interval 3: \( [3, 4] \)
\( s(4) = 2 + \dfrac{64}{3} – 32 + 12 = 2 + \dfrac{64}{3} – 20 = -18 + \dfrac{64}{3} = -\dfrac{54}{3} + \dfrac{64}{3} = \dfrac{10}{3} \)
Displacement: \( \dfrac{10}{3} – 2 = \dfrac{4}{3} \) (positive)
Total Distance = \( \dfrac{4}{3} + \dfrac{4}{3} + \dfrac{4}{3} = 4 \ \text{meters} \)