Math Problem Statement

Two cars, A and B, travel in a straight line. The distance of car A from the starting point is given as a function of time by xA(t) = αt + βt^2, with α = 2.60 m/s and β = 1.20 m/s^2. The distance of car B from the starting point is xB(t) = γt^2 - δt^3, with γ = 2.80 m/s^2 and δ = 0.20 m/s^3.

Solution

We are given the equations for the positions of two cars, AA and BB, as functions of time, with the following parameters:

For Car A: xA(t)=αt+βt2x_A(t) = \alpha t + \beta t^2 where α=2.60m/s\alpha = 2.60 \, \text{m/s} and β=1.20m/s2\beta = 1.20 \, \text{m/s}^2.

For Car B: xB(t)=γt2δt3x_B(t) = \gamma t^2 - \delta t^3 where γ=2.80m/s2\gamma = 2.80 \, \text{m/s}^2 and δ=0.20m/s3\delta = 0.20 \, \text{m/s}^3.

To understand their motion and how they relate to one another, we can analyze several aspects of their behavior:

1. Position of Each Car Over Time

  • For Car A: xA(t)=2.60t+1.20t2x_A(t) = 2.60t + 1.20t^2
  • For Car B: xB(t)=2.80t20.20t3x_B(t) = 2.80t^2 - 0.20t^3

2. Velocities

We can differentiate the position functions with respect to time tt to find the velocities.

  • For Car A, the velocity vA(t)v_A(t) is the derivative of xA(t)x_A(t): vA(t)=ddt(2.60t+1.20t2)=2.60+2×1.20t=2.60+2.40tv_A(t) = \frac{d}{dt} \left( 2.60t + 1.20t^2 \right) = 2.60 + 2 \times 1.20 t = 2.60 + 2.40t So, the velocity of Car A increases linearly with time.

  • For Car B, the velocity vB(t)v_B(t) is the derivative of xB(t)x_B(t): vB(t)=ddt(2.80t20.20t3)=2×2.80t3×0.20t2=5.60t0.60t2v_B(t) = \frac{d}{dt} \left( 2.80t^2 - 0.20t^3 \right) = 2 \times 2.80t - 3 \times 0.20t^2 = 5.60t - 0.60t^2 Car B's velocity is more complex: it initially increases, reaches a peak, and then decreases due to the negative cubic term.

3. When Do the Cars Have the Same Position?

To find the time tt at which the cars are at the same position, we set xA(t)=xB(t)x_A(t) = x_B(t): 2.60t+1.20t2=2.80t20.20t32.60t + 1.20t^2 = 2.80t^2 - 0.20t^3 This is a cubic equation, and we can solve it to find the time(s) when the cars are at the same position.

4. When Do the Cars Have the Same Velocity?

To find when the cars have the same velocity, we set vA(t)=vB(t)v_A(t) = v_B(t): 2.60+2.40t=5.60t0.60t22.60 + 2.40t = 5.60t - 0.60t^2 This is a quadratic equation and can be solved for tt to find when their velocities match.

Would you like me to solve these equations for you and provide the specific time values?


Five Related Questions:

  1. How do we find the acceleration of each car at any given time?
  2. What is the maximum speed Car B can achieve?
  3. At what time will Car A have traveled the same distance as Car B?
  4. How far from the starting point is each car after a specific time, say t=5secondst = 5 \, \text{seconds}?
  5. How can we calculate the relative velocity between Car A and Car B at a given time?

Tip:

When comparing two moving objects, analyzing their velocity and acceleration helps predict whether they will converge, diverge, or maintain a constant distance apart over time.

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Math Problem Analysis

Mathematical Concepts

Algebra
Differentiation
Cubic Equations
Quadratic Equations
Kinematics

Formulas

Position of Car A: x_A(t) = αt + βt^2
Position of Car B: x_B(t) = γt^2 - δt^3
Velocity of Car A: v_A(t) = 2.60 + 2.40t
Velocity of Car B: v_B(t) = 5.60t - 0.60t^2

Theorems

Basic Differentiation Rules
Solving Cubic and Quadratic Equations

Suitable Grade Level

Grades 10-12