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Let's solve the problem step by step.

### Problem Analysis
1. Two cars are moving towards each other. At \( t = 0 \), the distance between them is 1 km (1000 m).
2. **Car 1** is initially at rest but begins to accelerate at \( 3 \, \text{m/s}^2 \) from \( t = 0 \).
3. **Car 2** is moving at a constant speed of 40 km/h (which converts to \( 11.11 \, \text{m/s} \)) towards Car 1.

### Part (a): Time of Collision
To find the time of collision, we need to determine when the positions of Car 1 and Car 2 are the same.

Let:
- \( x_1(t) \) be the position of Car 1 at time \( t \).
- \( x_2(t) \) be the position of Car 2 at time \( t \).

**Car 1 Position:**
Since Car 1 starts from rest and accelerates with \( 3 \, \text{m/s}^2 \), its position at time \( t \) is given by:
\[
x_1(t) = \frac{1}{2} \cdot 3 \cdot t^2 = 1.5t^2 \, \text{m}
\]

**Car 2 Position:**
Car 2 starts at 1000 m and moves towards Car 1 at a constant speed of \( 11.11 \, \text{m/s} \), so its position at time \( t \) is:
\[
x_2(t) = 1000 - 11.11t \, \text{m}
\]

**Collision Condition:**
Set the positions equal to find the collision time:
\[
1.5t^2 = 1000 - 11.11t
\]

Rearrange into a standard quadratic form:
\[
1.5t^2 + 11.11t - 1000 = 0
\]

### Solving the Quadratic Equation
Let's solve this equation for \( t \) using the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1.5 \), \( b = 11.11 \), and \( c = -1000 \).

Calculating the values:

### Part (b): Speed of Car 1 Just Before Collision
Car 1's speed at time \( t \) can be found using the equation:
\[
v_1(t) = 3t
\]
We'll substitute the time of collision found in part (a) into this equation to find the speed.

### Part (c): Sketch the Graphs
For the graphs:
- **Position vs. Time:** Plot \( x_1(t) = 1.5t^2 \) and \( x_2(t) = 1000 - 11.11t \).
- **Velocity vs. Time:** Plot \( v_1(t) = 3t \) and a constant line at \( v_2(t) = 11.11 \).

Let me calculate the exact time of collision and the speed of Car 1 just before the collision.### Results:
1. **Time of Collision:** The two cars collide at approximately \( t = 22.38 \) seconds.
2. **Speed of Car 1 Just Before Collision:** The speed of Car 1 just before the collision is approximately \( 67.14 \, \text{m/s} \).

### Next Steps:
For part (c), you would sketch:
- **Position vs. Time Graphs:** Showing \( x_1(t) = 1.5t^2 \) for Car 1 and \( x_2(t) = 1000 - 11.11t \) for Car 2.
- **Velocity vs. Time Graphs:** Showing \( v_1(t) = 3t \) for Car 1 and a constant line at \( v_2 = 11.11 \, \text{m/s} \) for Car 2.

Would you like assistance with plotting these graphs or further details on the calculations?

### Related Questions:
1. How does the acceleration of Car 1 affect the collision time?
2. What would happen if Car 2 also had an acceleration instead of a constant speed?
3. How would the collision time change if the initial distance was different?
4. What would the speed of Car 1 be if it had a different acceleration?
5. How does the shape of the position vs. time graph differ between constant speed and acceleration?

**Tip:** When solving motion problems, always keep track of units (e.g., converting km/h to m/s) to avoid calculation errors.

Two cars drive directly toward each other. At time t = 0 the two cars are 1 km apart, car 1 is at rest, and car 2 is moving to the left with a speed 40 km/h. Car 1 begins to move at t = 0, speeding up with a constant acceleration of 3 m/s². Car 2 continues to move with a constant velocity. Find the time of collision, speed of car 1 before the collision, and sketch position vs. time and velocity vs. time graphs.

This physics problem calculates the time when two cars collide, one accelerating at 3 m/s² and the other moving at a constant 40 km/h. It involves solving quadratic equations to find the collision time and calculating the speed of the accelerating car before the collision. Suitable for high school students in Grades 9-12.

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