Math Problem Statement

If the position of a particle is s(t) = t^3/3 - 3t^2 + 4, find the velocity and position of the particle when its acceleration is zero.

Solution

The position function of the particle is given as:

s(t)=t33−3t2+4s(t) = \frac{t^3}{3} - 3t^2 + 4

We are asked to find the velocity and position of the particle when its acceleration is zero.

Step 1: Find the velocity function

The velocity v(t)v(t) is the derivative of the position function s(t)s(t) with respect to time tt:

v(t)=ds(t)dtv(t) = \frac{ds(t)}{dt}

Differentiating s(t)s(t):

= t^2 - 6t$$ ### Step 2: Find the acceleration function The acceleration $$a(t)$$ is the derivative of the velocity function $$v(t)$$: $$a(t) = \frac{dv(t)}{dt}$$ Differentiating $$v(t)$$: $$a(t) = 2t - 6$$ ### Step 3: Solve for when the acceleration is zero We need to find when $$a(t) = 0$$: $$2t - 6 = 0$$ Solving for $$t$$: $$t = 3$$ ### Step 4: Find the velocity when $$t = 3$$ Substitute $$t = 3$$ into the velocity function $$v(t)$$: $$v(3) = 3^2 - 6(3) = 9 - 18 = -9$$ Thus, the velocity when the acceleration is zero is $$v(3) = -9$$. ### Step 5: Find the position when $$t = 3$$ Substitute $$t = 3$$ into the position function $$s(t)$$: $$s(3) = \frac{3^3}{3} - 3(3^2) + 4 = \frac{27}{3} - 3(9) + 4 = 9 - 27 + 4 = -14$$ Thus, the position when the acceleration is zero is $$s(3) = -14$$. ### Final Answer: - The velocity of the particle when the acceleration is zero is $$v(3) = -9$$. - The position of the particle when the acceleration is zero is $$s(3) = -14$$. Would you like further details or have any questions? #### 5 Related Questions: 1. How do we interpret negative velocity in this context? 2. What does it mean physically for the acceleration to be zero at a specific time? 3. How would the graph of the position function $$s(t)$$ look? 4. What if the position function had more terms, like $$t^4$$? How would the process change? 5. Can we use these techniques to find the maximum or minimum position of the particle? #### Tip: Always check the units of each physical quantity (position, velocity, and acceleration) to ensure consistency when solving real-world problems.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Position, Velocity, and Acceleration

Formulas

v(t) = ds(t)/dt
a(t) = dv(t)/dt
v(t) = t^2 - 6t
a(t) = 2t - 6

Theorems

Differentiation

Suitable Grade Level

Grade 11-12 or early college (Calculus)

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