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.

We are given the equations for the positions of two cars, \( A \) and \( B \), as functions of time, with the following parameters:

For Car A:
\[
x_A(t) = \alpha t + \beta t^2
\]
where \( \alpha = 2.60 \, \text{m/s} \) and \( \beta = 1.20 \, \text{m/s}^2 \).

For Car B:
\[
x_B(t) = \gamma t^2 - \delta t^3
\]
where \( \gamma = 2.80 \, \text{m/s}^2 \) and \( \delta = 0.20 \, \text{m/s}^3 \).

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: 
  \[
  x_A(t) = 2.60t + 1.20t^2
  \]
- For Car B: 
  \[
  x_B(t) = 2.80t^2 - 0.20t^3
  \]

### 2. **Velocities**
We can differentiate the position functions with respect to time \( t \) to find the velocities.

- For Car A, the velocity \( v_A(t) \) is the derivative of \( x_A(t) \):
  \[
  v_A(t) = \frac{d}{dt} \left( 2.60t + 1.20t^2 \right) = 2.60 + 2 \times 1.20 t = 2.60 + 2.40t
  \]
  So, the velocity of Car A increases linearly with time.

- For Car B, the velocity \( v_B(t) \) is the derivative of \( x_B(t) \):
  \[
  v_B(t) = \frac{d}{dt} \left( 2.80t^2 - 0.20t^3 \right) = 2 \times 2.80t - 3 \times 0.20t^2 = 5.60t - 0.60t^2
  \]
  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 \( t \) at which the cars are at the same position, we set \( x_A(t) = x_B(t) \):
\[
2.60t + 1.20t^2 = 2.80t^2 - 0.20t^3
\]
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 \( v_A(t) = v_B(t) \):
\[
2.60 + 2.40t = 5.60t - 0.60t^2
\]
This is a quadratic equation and can be solved for \( t \) to find when their velocities match.

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

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### 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 \( t = 5 \, \text{seconds} \)?
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.

Explore the kinematic equations governing the motion of two cars, A and B, traveling in a straight line. The distance functions of both cars involve quadratic and cubic equations, and this problem includes the calculation of velocities using differentiation. Ideal for students in Grades 10-12.

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