Math Problem Statement

At time t, the position of a body moving along the s-axis is s = t^3 - 9t^2 + 24t m. (a) Find the body's acceleration each time the velocity is zero. (b) Find the body's speed each time the acceleration is zero. (c) Find the total distance traveled by the body from t = 0 to t = 3.

Solution

Let's solve each part step by step.

Given position function:

s(t)=t39t2+24ts(t) = t^3 - 9t^2 + 24t

(a) Find the body's acceleration each time the velocity is zero.

  1. Find velocity by differentiating s(t)s(t): v(t)=dsdt=3t218t+24v(t) = \frac{ds}{dt} = 3t^2 - 18t + 24

  2. Find when velocity is zero: 3t218t+24=03t^2 - 18t + 24 = 0 Divide by 3: t26t+8=0t^2 - 6t + 8 = 0 Factor: (t4)(t2)=0(t - 4)(t - 2) = 0 So, t=2t = 2 and t=4t = 4.

  3. Find acceleration by differentiating v(t)v(t): a(t)=dvdt=6t18a(t) = \frac{dv}{dt} = 6t - 18

  4. Evaluate acceleration at t=2t = 2 and t=4t = 4: a(2)=6(2)18=1218=6a(2) = 6(2) - 18 = 12 - 18 = -6 a(4)=6(4)18=2418=6a(4) = 6(4) - 18 = 24 - 18 = 6

    Answer: Acceleration is 6-6 m/s² at t=2t = 2 and 66 m/s² at t=4t = 4.


(b) Find the body's speed each time the acceleration is zero.

  1. Set acceleration to zero: 6t18=06t - 18 = 0 t=3t = 3

  2. Find speed at t=3t = 3 (speed is the absolute value of velocity): v(3)=3(3)218(3)+24v(3) = 3(3)^2 - 18(3) + 24 =3(9)54+24= 3(9) - 54 + 24 =2754+24=3= 27 - 54 + 24 = -3 Speed = 3=3| -3 | = 3 m/s.

    Answer: The body's speed when acceleration is zero is 33 m/s.


(c) Find the total distance traveled from t=0t = 0 to t=3t = 3.

  1. Find when the body changes direction (when velocity is zero in [0,3]):

    • We found v(t)=0v(t) = 0 at t=2t = 2, which is in [0,3].
  2. Find positions at t=0t = 0, t=2t = 2, and t=3t = 3: s(0)=(0)39(0)2+24(0)=0s(0) = (0)^3 - 9(0)^2 + 24(0) = 0 s(2)=(2)39(2)2+24(2)s(2) = (2)^3 - 9(2)^2 + 24(2) =836+48=20= 8 - 36 + 48 = 20 s(3)=(3)39(3)2+24(3)s(3) = (3)^3 - 9(3)^2 + 24(3) =2781+72=18= 27 - 81 + 72 = 18

  3. Compute total distance:

    • From t=0t = 0 to t=2t = 2: s(2)s(0)=200=20|s(2) - s(0)| = |20 - 0| = 20
    • From t=2t = 2 to t=3t = 3: s(3)s(2)=1820=2|s(3) - s(2)| = |18 - 20| = 2
    • Total distance = 20+2=2220 + 2 = 22 meters.

    Answer: The total distance traveled is 2222 meters.


Summary of Answers:

  • (a) Acceleration when velocity is zero: 6-6 m/s² at t=2t = 2, 66 m/s² at t=4t = 4.
  • (b) Speed when acceleration is zero: 33 m/s.
  • (c) Total distance traveled: 2222 meters.

Would you like a more detailed explanation for any part?


Related Questions:

  1. What is the velocity at t=4t = 4?
  2. How do we determine when the particle is speeding up or slowing down?
  3. What is the significance of the acceleration sign in motion analysis?
  4. Can we determine the average velocity from t=0t = 0 to t=3t = 3?
  5. How would the motion change if an additional force acted on the body?

Tip:

To analyze motion problems effectively, always differentiate the position function to get velocity and acceleration before solving! 🚀

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Math Problem Analysis

Mathematical Concepts

Differentiation
Kinematics
Polynomial Equations
Absolute Value

Formulas

Velocity: v(t) = ds/dt
Acceleration: a(t) = dv/dt
Quadratic Equation: ax^2 + bx + c = 0
Distance = |s(t2) - s(t1)|

Theorems

Roots of a quadratic equation
Relationship between velocity and acceleration
Absolute value in distance calculations

Suitable Grade Level

Grades 10-12