Kinematics Problem: Position and Velocity of a Particle

Math Problem Statement

Solve

A particle moves along the x-axis. Its position is given by the equation x = 2 + 3t - 4t^2 with x in meters and t in seconds. Determine (a) its position when it changes direction and (b) its velocity when it returns to the position it had at t = 0.

Solution

Let's solve the problem step by step.

Given:

The position of a particle is given by: $x(t) = 2 + 3t - 4t^2$ where $x$ is in meters and $t$ is in seconds.

We are tasked to:

Determine the position when the particle changes direction.
Determine the velocity when it returns to the position it had at $t = 0$ .

Part (a): Position when the particle changes direction

A particle changes direction when its velocity is zero. The velocity is the derivative of the position function: $v(t) = \frac{dx(t)}{dt}.$

Taking the derivative of $x(t)$ : $v(t) = \frac{d}{dt} \big(2 + 3t - 4t^2\big) = 3 - 8t.$

Set $v(t) = 0$ to find when the particle changes direction: $3 - 8t = 0$ $t = \frac{3}{8} \, \text{seconds}.$

Now substitute $t = \frac{3}{8}$ into the position function $x(t)$ to find the position: $x\left(\frac{3}{8}\right) = 2 + 3\left(\frac{3}{8}\right) - 4\left(\frac{3}{8}\right)^2.$

Simplify:

= 2 + \frac{9}{8} - \frac{36}{64}.$$ Convert everything to a common denominator of 64: $$x\left(\frac{3}{8}\right) = \frac{128}{64} + \frac{72}{64} - \frac{36}{64}.$$ Simplify: $$x\left(\frac{3}{8}\right) = \frac{164}{64} = \frac{41}{16} \, \text{meters}.$$ --- ### **Part (b): Velocity when the particle returns to $$x(t=0)$$** First, find $$x(0)$$: $$x(0) = 2 + 3(0) - 4(0)^2 = 2 \, \text{meters}.$$ Now find the times when the particle is at $$x = 2$$. Set $$x(t) = 2$$: $$2 + 3t - 4t^2 = 2.$$ Simplify: $$3t - 4t^2 = 0.$$ Factorize: $$t(3 - 4t) = 0.$$ Solve: $$t = 0 \quad \text{or} \quad t = \frac{3}{4} \, \text{seconds}.$$ The particle is at $$x = 2$$ at $$t = 0$$ (initial position) and $$t = \frac{3}{4}$$. Find the velocity at $$t = \frac{3}{4}$$: $$v(t) = 3 - 8t.$$ Substitute $$t = \frac{3}{4}$$: $$v\left(\frac{3}{4}\right) = 3 - 8\left(\frac{3}{4}\right) = 3 - 6 = -3 \, \text{m/s}.$$ --- ### **Final Answers:** 1. The position when the particle changes direction is: $$x = \frac{41}{16} \, \text{meters}.$$ 2. The velocity when the particle returns to $$x = 2$$ is: $$v = -3 \, \text{m/s}.$$ --- Do you want further details or have any questions? --- ### **Related Questions:** 1. What is the acceleration of the particle at any given time? 2. At what time does the particle reach its maximum or minimum position? 3. What is the total displacement of the particle between $$t = 0$$ and $$t = 1$$? 4. Can you calculate the particle's average velocity over a given time interval? 5. How would the motion of the particle change if the equation had an additional cubic term? ### **Tip:** When solving kinematics problems, always start by identifying key points (e.g., zero velocity for direction changes) and take derivatives carefully for velocity and acceleration.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Kinematics
Quadratic Equations

Formulas

Velocity: v(t) = dx(t)/dt
Quadratic formula: ax^2 + bx + c = 0

Theorems

Zero derivative theorem for direction change
Factorization for solving quadratic equations

Suitable Grade Level

Grades 10-12

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