Math Problem Statement

A particle moves along the y-axis so that at time t ≥ 0 its position is given by y(t) = t³ - 6t² + 9t. Over the time interval 0 < t < 4, for what values of t is the speed of the particle increasing?

Solution

This question asks about when the speed of a particle moving along the y-axis is increasing. The position of the particle is given as y(t)=t36t2+9ty(t) = t^3 - 6t^2 + 9t. To determine when the speed is increasing, we follow these steps:


1. Determine Velocity and Acceleration

The speed of the particle is increasing when the velocity and acceleration have the same sign.

  • Velocity (v(t)v(t)) is the first derivative of position: v(t)=dydt=3t212t+9v(t) = \frac{dy}{dt} = 3t^2 - 12t + 9

  • Acceleration (a(t)a(t)) is the derivative of velocity: a(t)=dvdt=6t12a(t) = \frac{dv}{dt} = 6t - 12


2. Find Critical Points

  • Set v(t)=0v(t) = 0 to find critical points for velocity: 3t212t+9=0    t24t+3=0    (t3)(t1)=03t^2 - 12t + 9 = 0 \implies t^2 - 4t + 3 = 0 \implies (t-3)(t-1) = 0 So, t=1t = 1 and t=3t = 3.

  • Set a(t)=0a(t) = 0 to find critical points for acceleration: 6t12=0    t=26t - 12 = 0 \implies t = 2


3. Analyze Signs of Velocity and Acceleration

To determine when the speed is increasing, evaluate the signs of v(t)v(t) and a(t)a(t) over the intervals defined by t=1t = 1, t=2t = 2, and t=3t = 3:

Intervals: 0<t<10 < t < 1, 1<t<21 < t < 2, 2<t<32 < t < 3, 3<t<43 < t < 4

  • For v(t)v(t):

    • Factor v(t)=3(t1)(t3)v(t) = 3(t-1)(t-3):
      • 0<t<10 < t < 1: v(t)>0v(t) > 0
      • 1<t<31 < t < 3: v(t)<0v(t) < 0
      • t>3t > 3: v(t)>0v(t) > 0
  • For a(t)a(t):

    • a(t)=6(t2)a(t) = 6(t-2):
      • 0<t<20 < t < 2: a(t)<0a(t) < 0
      • t>2t > 2: a(t)>0a(t) > 0

Speed Increasing:

The speed increases when v(t)v(t) and a(t)a(t) have the same sign:

  • 1<t<21 < t < 2: Both v(t)<0v(t) < 0 and a(t)<0a(t) < 0.
  • 3<t<43 < t < 4: Both v(t)>0v(t) > 0 and a(t)>0a(t) > 0.

4. Answer

The correct intervals are 1<t<21 < t < 2 and 3<t<43 < t < 4. Therefore, the answer is:

D: 1<t<2 and 3<t<4\boxed{\text{D: } 1 < t < 2 \text{ and } 3 < t < 4}


Would you like a step-by-step explanation for any part?

Related Questions:

  1. How do you determine the critical points for a given polynomial?
  2. Why do the signs of velocity and acceleration determine when speed is increasing?
  3. How can you graphically represent v(t)v(t) and a(t)a(t)?
  4. Can speed ever decrease if v(t)v(t) and a(t)a(t) have opposite signs?
  5. How does this problem extend to higher dimensions of motion?

Tip:

Always check the relationship between velocity and acceleration when analyzing the change in speed.

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Math Problem Analysis

Mathematical Concepts

Calculus
Velocity
Acceleration
Speed

Formulas

Velocity v(t) = dy/dt
Acceleration a(t) = dv/dt
Critical points are found by setting the derivatives to zero

Theorems

Sign of velocity and acceleration determines speed behavior

Suitable Grade Level

Grades 11-12 (AP Calculus or equivalent)