Math Problem Statement

A particle moves along the x-axis so that its position at time t is given by x(t) = t^2 - 6t + 5. For what value of t is the velocity of the particle zero?

Solution

To find the time tt at which the velocity of the particle is zero, we start by finding the derivative of the position function x(t)x(t), as this derivative represents the velocity function v(t)v(t).

Given: x(t)=t26t+5x(t) = t^2 - 6t + 5

  1. Find the velocity function v(t)v(t): v(t)=dxdt=ddt(t26t+5)=2t6v(t) = \frac{dx}{dt} = \frac{d}{dt}(t^2 - 6t + 5) = 2t - 6

  2. Set the velocity to zero and solve for tt: 2t6=02t - 6 = 0 2t=62t = 6 t=3t = 3

So, the velocity of the particle is zero at t=3t = 3.

Would you like a detailed explanation of each step?

Here are five related questions to deepen your understanding:

  1. What is the significance of setting the velocity to zero in this context?
  2. How can we determine if the particle changes direction when the velocity is zero?
  3. What would the acceleration function of the particle be, and how can it help us understand its motion?
  4. If x(t)x(t) represented the position in three-dimensional space, how would we approach this problem differently?
  5. How can we find the time intervals during which the particle moves in a positive or negative direction?

Tip: In physics problems involving motion, setting the derivative of the position to zero often reveals critical points in the motion, such as stops or changes in direction.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Velocity

Formulas

Velocity as the derivative of position: v(t) = dx/dt

Theorems

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Suitable Grade Level

Grades 11-12