Math Problem Statement

The position of a particle in the xy-plane at time t is r(t) = (t - 4)i + (t^2 + 1)j. Find an equation in x and y whose graph is the path of the particle. Then find the particle's velocity and acceleration vectors at t = 5.

Solution

We are given the position of a particle in the xy-plane as:

r(t)=(t4)i^+(t2+1)j^r(t) = (t - 4)\hat{i} + (t^2 + 1)\hat{j}

Our goal is to find:

  1. The equation for the path of the particle in terms of yy and xx.
  2. The velocity vector at t=5t = 5.
  3. The acceleration vector at t=5t = 5.

Step 1: Equation for the path of the particle

We are given:

  • x(t)=t4x(t) = t - 4
  • y(t)=t2+1y(t) = t^2 + 1

To eliminate tt, solve for tt in terms of xx: x=t4    t=x+4x = t - 4 \implies t = x + 4

Substitute t=x+4t = x + 4 into the equation for yy: y=(x+4)2+1y = (x + 4)^2 + 1 Simplify: y=x2+8x+16+1    y=x2+8x+17y = x^2 + 8x + 16 + 1 \implies y = x^2 + 8x + 17

Thus, the equation for the path is: y=x2+8x+17y = x^2 + 8x + 17

Step 2: Velocity vector at t=5t = 5

The velocity vector is the derivative of the position vector: v(t)=ddt[(t4)i^+(t2+1)j^]v(t) = \frac{d}{dt}[(t - 4)\hat{i} + (t^2 + 1)\hat{j}] Differentiate each component: v(t)=(1i^+2tj^)v(t) = \left(1\hat{i} + 2t\hat{j}\right)

At t=5t = 5: v(5)=1i^+2(5)j^=1i^+10j^v(5) = 1\hat{i} + 2(5)\hat{j} = 1\hat{i} + 10\hat{j}

Thus, the velocity vector at t=5t = 5 is: v=1i^+10j^v = 1\hat{i} + 10\hat{j}

Step 3: Acceleration vector at t=5t = 5

The acceleration vector is the derivative of the velocity vector: a(t)=ddt[1i^+2tj^]=0i^+2j^a(t) = \frac{d}{dt}[1\hat{i} + 2t\hat{j}] = 0\hat{i} + 2\hat{j}

At t=5t = 5: a(5)=0i^+2j^a(5) = 0\hat{i} + 2\hat{j}

Thus, the acceleration vector at t=5t = 5 is: a=0i^+2j^a = 0\hat{i} + 2\hat{j}

Final Answers:

  • The equation for the path is: y=x2+8x+17y = x^2 + 8x + 17.
  • The velocity vector at t=5t = 5 is: 1i^+10j^1\hat{i} + 10\hat{j}.
  • The acceleration vector at t=5t = 5 is: 0i^+2j^0\hat{i} + 2\hat{j}.

Would you like further details on any step?

Here are 5 related questions:

  1. How do we derive the velocity and acceleration vectors from the position vector in 3D?
  2. What does the equation of the path tell us about the shape of the particle's trajectory?
  3. How would the velocity and acceleration vectors change for different values of tt?
  4. How do velocity and acceleration vectors relate to the speed and direction of the particle?
  5. Can we determine the particle's speed at t=5t = 5?

Tip: The velocity vector provides both the speed and direction of motion. You can calculate speed by finding the magnitude of the velocity vector.

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Parametric Equations
Derivatives
Kinematics

Formulas

Velocity vector: v(t) = dr/dt
Acceleration vector: a(t) = dv/dt
Path equation: y = f(x)

Theorems

Derivative rules for vector functions

Suitable Grade Level

Grades 11-12