To solve this problem, we will use the hints provided and set up the equations for the positions of both the car and the motorcycle. Then, we'll solve for the time it takes for the motorcycle to catch up with the car.

Step-by-Step Solution

1. Set Up the Equations of Motion

• Car's Position Equation: The car is traveling at a constant speed, so its position $x_c(t)$ at time $t$ after $t_1 = 4.00 \, \text{s}$ can be described by:

$x_c(t) = x_{1,c} + v_{1,c} (t - t_1) + \frac{1}{2} a_c (t - t_1)^2$

Given that $x_{1,c} = 94.0 \, \text{m}$ (the initial position of the car relative to the motorcycle), $v_{1,c} = 22.5 \, \text{m/s}$, and $a_c = 0 \, \text{m/s}^2$ (since the car is moving at a constant speed), the equation simplifies to:

$x_c(t) = 94.0 + 22.5 (t - 4.00)$

• Motorcycle's Position Equation: The motorcycle starts accelerating at $t_1 = 4.00 \, \text{s}$, so its position $x_m(t)$ at time $t \geq t_1$ is:

$x_m(t) = x_{1,m} + v_{1,m} (t - t_1) + \frac{1}{2} a_m (t - t_1)^2$

Here, $x_{1,m} = 0$ (as we set the initial position of the motorcycle to be zero), $v_{1,m} = 22.5 \, \text{m/s}$, and $a_m = 5.00 \, \text{m/s}^2$. Thus, the equation becomes:

$x_m(t) = 0 + 22.5 (t - 4.00) + \frac{1}{2} \cdot 5.00 (t - 4.00)^2$

Simplifying, we have:

$x_m(t) = 22.5 (t - 4.00) + 2.5 (t - 4.00)^2$

2. Set the Positions Equal to Each Other to Find $t_2 - t_1$

At the time $t = t_2$, the motorcycle catches up with the car, meaning $x_c(t_2) = x_m(t_2)$. Therefore, we set the two equations equal to each other:

$94.0 + 22.5 (t - 4.00) = 22.5 (t - 4.00) + 2.5 (t - 4.00)^2$

Simplifying, we can cancel the linear terms $22.5 (t - 4.00)$ on both sides:

$94.0 = 2.5 (t - 4.00)^2$

3. Solve for $t_2 - t_1$

Now solve for $(t - 4.00)$:

$(t - 4.00)^2 = \frac{94.0}{2.5}$

$(t - 4.00)^2 = 37.6$

Take the square root of both sides:

$t - 4.00 = \sqrt{37.6}$

$t - 4.00 \approx 6.13$

Thus:

$t_2 - t_1 = t - 4.00 = 6.13 \, \text{s}$

Final Answer

The time $t_2 - t_1$ it takes from the moment the motorcycle starts to accelerate until it catches up with the car is 6.13 seconds (to three significant figures).

Would you like more details or have any questions?

Here are 5 related questions to consider:

1. What if the initial distance between the car and motorcycle was different? How would that affect $t_2 - t_1$?
2. How would the result change if the motorcycle had a different acceleration?
3. How far does the motorcycle travel while catching up to the car?
4. What would happen if the car also started accelerating after $t_1$?
5. Can you calculate the relative speed of the motorcycle as it catches up with the car?

Tip: When working with relative motion problems, remember to define a consistent frame of reference and account for initial conditions accurately.