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 21.0 m/s
 , and the distance between them is 95.0 m
 . After t1
 = 3.00 s
 , the motorcycle starts to accelerate at a rate of 7.00 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
 .

To solve the problem, we need to find how long it takes for the motorcycle to catch up with the car after it starts accelerating. Here's how we can approach the solution:

### Given Data:
- Initial speed of both the car and the motorcycle, \( v_0 = 21.0 \, \text{m/s} \)
- Initial distance between them, \( d = 95.0 \, \text{m} \)
- Motorcycle starts accelerating after \( t_1 = 3.00 \, \text{s} \)
- Acceleration of the motorcycle, \( a = 7.00 \, \text{m/s}^2 \)
- The car continues at constant speed.

We need to find the time \( t_2 - t_1 \) (i.e., the time after \( t_1 \) until the motorcycle catches up with the car).

### Step 1: Motion of the Car
Since the car is moving at a constant speed, the position of the car as a function of time after \( t_1 \) is:

\[
x_{\text{car}}(t) = v_0 \cdot t
\]
where \( t \) is the time after \( t_1 \), so \( t = 0 \) at \( t_1 \).

### Step 2: Motion of the Motorcycle
The motorcycle starts accelerating at \( t_1 \). Its motion can be described by the equation for uniformly accelerated motion:

\[
x_{\text{motorcycle}}(t) = v_0 \cdot t + \frac{1}{2} a \cdot t^2
\]
where \( t \) is the time after \( t_1 \).

### Step 3: Setting Up the Equation
The motorcycle catches up with the car when their positions are equal. Initially, the motorcycle is behind by \( d = 95.0 \, \text{m} \). Therefore, we set up the equation:

\[
x_{\text{motorcycle}}(t) = x_{\text{car}}(t) + 95.0
\]

Substituting the equations for \( x_{\text{motorcycle}}(t) \) and \( x_{\text{car}}(t) \):

\[
v_0 \cdot t + \frac{1}{2} a \cdot t^2 = v_0 \cdot t + 95.0
\]

### Step 4: Simplifying the Equation
Cancelling out \( v_0 \cdot t \) from both sides:

\[
\frac{1}{2} a \cdot t^2 = 95.0
\]

Now, substitute \( a = 7.00 \, \text{m/s}^2 \):

\[
\frac{1}{2} \cdot 7.00 \cdot t^2 = 95.0
\]

This simplifies to:

\[
3.50 \cdot t^2 = 95.0
\]

### Step 5: Solving for \( t \)
Solve for \( t^2 \):

\[
t^2 = \frac{95.0}{3.50}
\]

\[
t^2 = 27.14
\]

Now, take the square root of both sides:

\[
t = \sqrt{27.14} \approx 5.21 \, \text{seconds}
\]

### Step 6: Conclusion
The time \( t_2 - t_1 \) (the time it takes for the motorcycle to catch up) is approximately \( 5.21 \, \text{seconds} \).

### Final Answer:
The motorcycle catches up with the car after approximately \( 5.21 \, \text{seconds} \) from the moment it starts accelerating.

Do you have any questions or would you like more details?

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Here are 5 related questions you might explore:
1. How far does the motorcycle travel during this time?
2. What is the final velocity of the motorcycle when it catches up with the car?
3. What would happen if the acceleration of the motorcycle was different?
4. How long would it take if the initial distance was doubled?
5. How does the relative motion of the two vehicles change if the car also accelerates?

**Tip:** Always pay attention to which quantities are constant and which are changing in kinematics problems—it helps simplify the equations!

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 21.0 m/s, and the distance between them is 95.0 m. After t1 = 3.00 s, the motorcycle starts to accelerate at a rate of 7.00 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.

Solve a kinematics problem where a motorcycle accelerates to catch up with a car traveling at a constant speed. Initial speed is 21.0 m/s, distance between them is 95.0 m, and the motorcycle accelerates at 7.00 m/s². Find how long it takes for the motorcycle to catch up after accelerating. Suitable for students in Grades 11-12.

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