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.

Which car is ahead just after they leave the starting point?

At what time(s) are the cars at the same point?

Let's tackle each part of this problem step by step.

### 1. Which car is ahead just after they leave the starting point?

To find out which car is ahead just after they leave the starting point (i.e., at time \( t = 0 \)), we need to evaluate the positions of car A and car B at \( t = 0 \).

The position functions of the cars are given as:

- Car A: \( x_A(t) = \alpha t + \beta t^2 \)
- Car B: \( x_B(t) = \gamma t^2 - \delta t^3 \)

First, substitute \( t = 0 \) into both position functions:

#### Car A at \( t = 0 \):
\[
x_A(0) = \alpha \cdot 0 + \beta \cdot 0^2 = 0
\]

#### Car B at \( t = 0 \):
\[
x_B(0) = \gamma \cdot 0^2 - \delta \cdot 0^3 = 0
\]

At \( t = 0 \), both cars are at the starting point, meaning neither car is ahead just as they leave the starting point.

### 2. At what time(s) are the cars at the same point?

The cars will be at the same point when their positions are equal, i.e., when \( x_A(t) = x_B(t) \).

The position functions are:
\[
\alpha t + \beta t^2 = \gamma t^2 - \delta t^3
\]
Substitute the known values of the constants \( \alpha = 2.60 \, \text{m/s} \), \( \beta = 1.20 \, \text{m/s}^2 \), \( \gamma = 2.80 \, \text{m/s}^2 \), and \( \delta = 0.20 \, \text{m/s}^3 \):
\[
2.60 t + 1.20 t^2 = 2.80 t^2 - 0.20 t^3
\]

Rearranging the equation:
\[
0.20 t^3 + (1.60 t^2) - (2.60 t) = 0
\]
This is a cubic equation:
\[
0.20 t^3 + 1.60 t^2 - 2.60 t = 0
\]
Factor out \( t \):
\[
t(0.20 t^2 + 1.60 t - 2.60) = 0
\]
One solution is \( t = 0 \), which means they are at the same point at \( t = 0 \) (as we already established).

Now, solve the quadratic equation:
\[
0.20 t^2 + 1.60 t - 2.60 = 0
\]
Dividing by 0.20 to simplify:
\[
t^2 + 8t - 13 = 0
\]
Now solve this using the quadratic formula:
\[
t = \frac{-8 \pm \sqrt{8^2 - 4(1)(-13)}}{2(1)}
\]
\[
t = \frac{-8 \pm \sqrt{64 + 52}}{2}
\]
\[
t = \frac{-8 \pm \sqrt{116}}{2}
\]
\[
t = \frac{-8 \pm 10.77}{2}
\]

This gives two solutions:
\[
t = \frac{-8 + 10.77}{2} \approx 1.39 \, \text{seconds}
\]
\[
t = \frac{-8 - 10.77}{2} \approx -9.39 \, \text{seconds (not physically meaningful)}
\]

Thus, the cars are at the same position again at \( t \approx 1.39 \) seconds after they start.

### Final Answers:
1. Just after they leave the starting point, both cars are at the same position (neither is ahead).
2. The cars are at the same position again at \( t \approx 1.39 \) seconds.

Would you like more details or have any questions about this? 

Here are 5 related questions to expand your understanding:
1. What are the speeds of both cars at \( t = 0 \)?
2. How does the acceleration of each car differ over time?
3. What is the distance traveled by each car at \( t = 1.39 \) seconds?
4. Can you find the relative velocity between the two cars at the point they meet again?
5. How would the solution change if the cubic term \( \delta t^3 \) for car B were different?

**Tip:** When solving quadratic or cubic equations, always check if factoring or simple numerical methods can simplify the equation before applying the quadratic formula!

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. Which car is ahead just after they leave the starting point? At what time(s) are the cars at the same point?

Explore how to solve a kinematics problem where two cars' positions are modeled by quadratic and cubic equations. Learn how to find when they are at the same point using algebra and the quadratic formula. Suitable for students in Grades 10-12.

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