A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle are both traveling at the same speed of 19.5 m/s
 , and the distance between them is 90.0 m
 . After t1
 = 3.00 s
 , the motorcycle starts to accelerate at a rate of 4.00 m/s2
 . The motorcycle catches up with the car at some time t2
How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car? In other words, find  t2−t1
 .

We need to find the time it takes for the motorcycle to catch up with the car after the motorcycle starts accelerating.

### Given:
- Initial speed of both the car and the motorcycle: \( v_0 = 19.5 \, \text{m/s} \)
- Initial distance between the car and the motorcycle: \( d_0 = 90.0 \, \text{m} \)
- The motorcycle starts accelerating after \( t_1 = 3.00 \, \text{s} \).
- Acceleration of the motorcycle: \( a = 4.00 \, \text{m/s}^2 \)

### To find:
- The time \( \Delta t = t_2 - t_1 \), where \( t_2 \) is the time when the motorcycle catches up with the car after the motorcycle starts accelerating.

### Step-by-step solution:

1. **Position of the car**:
   The car is moving at a constant speed, so its position at any time \( t \) after \( t_1 \) can be described by:
   \[
   x_{\text{car}}(t) = v_0 \cdot t
   \]

2. **Position of the motorcycle**:
   After the motorcycle starts accelerating at \( t = t_1 \), its position at any time \( t \) after it starts accelerating is:
   \[
   x_{\text{motorcycle}}(t) = v_0 \cdot t + \frac{1}{2} a \cdot (t - t_1)^2
   \]
   where \( a = 4.00 \, \text{m/s}^2 \) is the acceleration.

3. **Catching up condition**:
   The motorcycle catches up to the car when their positions are the same:
   \[
   x_{\text{motorcycle}}(t_2) = x_{\text{car}}(t_2)
   \]
   Substituting the expressions for \( x_{\text{car}}(t) \) and \( x_{\text{motorcycle}}(t) \) into this equation gives:
   \[
   v_0 \cdot t_2 + \frac{1}{2} a \cdot (t_2 - t_1)^2 = v_0 \cdot t_2
   \]

4. **Solve for \( t_2 - t_1 \)**:
   Since \( v_0 \cdot t_2 \) cancels out on both sides, we're left with:
   \[
   \frac{1}{2} a \cdot (t_2 - t_1)^2 = d_0
   \]
   Substituting the given values \( a = 4.00 \, \text{m/s}^2 \) and \( d_0 = 90.0 \, \text{m} \), we have:
   \[
   \frac{1}{2} \cdot 4.00 \cdot (t_2 - t_1)^2 = 90.0
   \]
   Simplifying:
   \[
   2.00 \cdot (t_2 - t_1)^2 = 90.0
   \]
   \[
   (t_2 - t_1)^2 = \frac{90.0}{2.00}
   \]
   \[
   (t_2 - t_1)^2 = 45.0
   \]
   Taking the square root:
   \[
   t_2 - t_1 = \sqrt{45.0} \approx 6.71 \, \text{seconds}
   \]

### Final answer:
The time it takes for the motorcycle to catch up with the car after it starts accelerating is approximately \( 6.71 \, \text{seconds} \).

Would you like more details or have any questions?

Here are some related questions you could explore:
1. What is the distance traveled by the motorcycle while catching up?
2. How fast is the motorcycle traveling when it catches up?
3. What if the motorcycle accelerated at a different rate?
4. How much total time does it take for the motorcycle to catch up, including the first 3 seconds?
5. What if the initial distance was different?

**Tip:** When analyzing motion problems, breaking them into phases (constant speed, acceleration, etc.) makes solving much easier!

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle are both traveling at the same speed of 19.5 m/s, and the distance between them is 90.0 m. After t1 = 3.00 s, the motorcycle starts to accelerate at a rate of 4.00 m/s². The motorcycle catches up with the car at some time t2. How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car?

Explore a step-by-step solution to a kinematics problem where a motorcycle accelerates to catch up with a car. This example covers key physics concepts such as constant speed, acceleration, and quadratic equations. Ideal for high school students in Grades 10-12.

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