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.