Math Problem Statement

A particle is moving with acceleration a(t) = 36t + 12. Its position at time t = 0 is s(0) = 3, and its velocity at time t = 0 is v(0) = 8. What is its position at time t = 5?

Solution

To solve for the particle's position at time t=5t = 5, given the acceleration, initial velocity, and initial position, we can follow these steps:

  1. Given Information:

    • Acceleration: a(t)=36t+12a(t) = 36t + 12
    • Initial Position: s(0)=3s(0) = 3
    • Initial Velocity: v(0)=8v(0) = 8
  2. Step 1: Integrate Acceleration to Find Velocity
    We start by integrating the acceleration function to get the velocity function.

    v(t)=a(t)dt=(36t+12)dtv(t) = \int a(t) \, dt = \int (36t + 12) \, dt

    This integration yields:

    v(t)=18t2+12t+Cv(t) = 18t^2 + 12t + C

    Using the initial condition v(0)=8v(0) = 8:

    v(0)=18(0)2+12(0)+C=8C=8v(0) = 18(0)^2 + 12(0) + C = 8 \Rightarrow C = 8

    Thus, the velocity function is:

    v(t)=18t2+12t+8v(t) = 18t^2 + 12t + 8

  3. Step 2: Integrate Velocity to Find Position
    Next, we integrate the velocity function to get the position function.

    s(t)=v(t)dt=(18t2+12t+8)dts(t) = \int v(t) \, dt = \int (18t^2 + 12t + 8) \, dt

    Performing this integration:

    s(t)=6t3+6t2+8t+Cs(t) = 6t^3 + 6t^2 + 8t + C

    Using the initial condition s(0)=3s(0) = 3:

    s(0)=6(0)3+6(0)2+8(0)+C=3C=3s(0) = 6(0)^3 + 6(0)^2 + 8(0) + C = 3 \Rightarrow C = 3

    Therefore, the position function is:

    s(t)=6t3+6t2+8t+3s(t) = 6t^3 + 6t^2 + 8t + 3

  4. Step 3: Substitute t=5t = 5 to Find s(5)s(5)

    Now we calculate s(5)s(5):

    s(5)=6(5)3+6(5)2+8(5)+3s(5) = 6(5)^3 + 6(5)^2 + 8(5) + 3

    Breaking it down:

    • 6(5)3=6125=7506(5)^3 = 6 \cdot 125 = 750
    • 6(5)2=625=1506(5)^2 = 6 \cdot 25 = 150
    • 85=408 \cdot 5 = 40
    • Summing these up: 750+150+40+3=943750 + 150 + 40 + 3 = 943

    Therefore, the position of the particle at t=5t = 5 is:

    s(5)=943s(5) = 943

Would you like further details or have any questions about this solution?

Here are five related questions to explore:

  1. How would the solution change if the initial velocity v(0)v(0) were different?
  2. What would be the particle's position function if the acceleration were a(t)=12t+6a(t) = 12t + 6 instead?
  3. How can we determine the particle's speed at t=5t = 5 using the velocity function derived?
  4. What if the initial position s(0)s(0) was negative—how would that impact s(5)s(5)?
  5. How does integrating acceleration give us velocity, and integrating velocity give us position?

Tip: When working with initial conditions, double-check each value after integration to ensure the constants are applied correctly.

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Math Problem Analysis

Mathematical Concepts

Calculus
Kinematics
Integration
Initial Value Problems

Formulas

Position s(t) from velocity function v(t)
Velocity v(t) from acceleration a(t)

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12