Math Problem Statement

Find the velocity, acceleration, and speed of a particle with the given position function. r(t) = t2i + 4tj + 6 ln(t)k

Solution

Given the position function r(t)=t2i+4tj+6ln(t)k\mathbf{r}(t) = t^2 \mathbf{i} + 4t \mathbf{j} + 6 \ln(t) \mathbf{k}, let's find the velocity, acceleration, and speed of the particle.

1. Velocity (v(t)\mathbf{v}(t))

The velocity vector is the first derivative of the position vector with respect to time tt: v(t)=ddtr(t)\mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) Differentiating each component of r(t)\mathbf{r}(t): v(t)=ddt(t2)i+ddt(4t)j+ddt(6ln(t))k\mathbf{v}(t) = \frac{d}{dt}(t^2) \mathbf{i} + \frac{d}{dt}(4t) \mathbf{j} + \frac{d}{dt}(6 \ln(t)) \mathbf{k} v(t)=2ti+4j+6tk\mathbf{v}(t) = 2t \mathbf{i} + 4 \mathbf{j} + \frac{6}{t} \mathbf{k} Thus, the velocity vector is: v(t)=2ti+4j+6tk\mathbf{v}(t) = 2t \mathbf{i} + 4 \mathbf{j} + \frac{6}{t} \mathbf{k}

2. Acceleration (a(t)\mathbf{a}(t))

The acceleration vector is the second derivative of the position vector with respect to time tt, or the first derivative of the velocity vector: a(t)=ddtv(t)\mathbf{a}(t) = \frac{d}{dt} \mathbf{v}(t) Differentiating each component of v(t)\mathbf{v}(t): a(t)=ddt(2t)i+ddt(4)j+ddt(6t)k\mathbf{a}(t) = \frac{d}{dt}(2t) \mathbf{i} + \frac{d}{dt}(4) \mathbf{j} + \frac{d}{dt}\left(\frac{6}{t}\right) \mathbf{k} a(t)=2i+0j6t2k\mathbf{a}(t) = 2 \mathbf{i} + 0 \mathbf{j} - \frac{6}{t^2} \mathbf{k} Thus, the acceleration vector is: a(t)=2i6t2k\mathbf{a}(t) = 2 \mathbf{i} - \frac{6}{t^2} \mathbf{k}

3. Speed (v(t)|\mathbf{v}(t)|)

The speed is the magnitude of the velocity vector: v(t)=(2t)2+(4)2+(6t)2|\mathbf{v}(t)| = \sqrt{(2t)^2 + (4)^2 + \left(\frac{6}{t}\right)^2} v(t)=4t2+16+36t2|\mathbf{v}(t)| = \sqrt{4t^2 + 16 + \frac{36}{t^2}} Thus, the speed is: v(t)=4t2+16+36t2|\mathbf{v}(t)| = \sqrt{4t^2 + 16 + \frac{36}{t^2}}

Summary:

  • Velocity: v(t)=2ti+4j+6tk\mathbf{v}(t) = 2t \mathbf{i} + 4 \mathbf{j} + \frac{6}{t} \mathbf{k}
  • Acceleration: a(t)=2i6t2k\mathbf{a}(t) = 2 \mathbf{i} - \frac{6}{t^2} \mathbf{k}
  • Speed: v(t)=4t2+16+36t2|\mathbf{v}(t)| = \sqrt{4t^2 + 16 + \frac{36}{t^2}}

Do you have any questions or would you like further clarification?

Here are 5 related questions you might find helpful:

  1. How do you find the unit vector of the velocity?
  2. What is the significance of speed versus velocity?
  3. How would you compute the distance traveled by the particle?
  4. What are the initial conditions for velocity and acceleration when t=1t = 1?
  5. How does the motion change as tt \to \infty?

Tip: To analyze a particle's motion, always start by finding the velocity and acceleration as functions of time to understand its dynamic behavior.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Vector Calculus
Derivatives
Kinematics

Formulas

Velocity: v(t) = dr(t)/dt
Acceleration: a(t) = dv(t)/dt
Speed: |v(t)| = sqrt((2t)^2 + (4)^2 + (6/t)^2)

Theorems

Chain Rule
Magnitude of a Vector

Suitable Grade Level

Undergraduate Level (Calculus II or Physics)