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 18.0 m/s
 , and the distance between them is 91.0 m
 . After t1
 = 4.00 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
 .
Express the time numerically in seconds using three significant figures.


Let's define the variables:

- Initial speed of both the car and motorcycle: \( v_0 = 18.0 \, \text{m/s} \)
- Distance between the car and motorcycle initially: \( d_0 = 91.0 \, \text{m} \)
- Time after which the motorcycle starts to accelerate: \( t_1 = 4.00 \, \text{s} \)
- Acceleration of the motorcycle: \( a = 7.00 \, \text{m/s}^2 \)

We are tasked with finding the time \( \Delta t = t_2 - t_1 \), which represents the duration it takes for the motorcycle to catch up with the car after it begins accelerating.

### Step 1: Distance traveled by the car
Since the car is traveling at a constant speed, the distance it covers after time \( t_1 \) is given by:

\[
d_{\text{car}} = v_0 \cdot \Delta t
\]

### Step 2: Distance traveled by the motorcycle
The motorcycle travels at a constant speed for \( t_1 = 4.00 \, \text{s} \), so the distance it covers during this period is:

\[
d_{\text{motorcycle1}} = v_0 \cdot t_1
\]

After \( t_1 \), the motorcycle accelerates, so the distance covered after it starts accelerating (from \( t_1 \) onwards) is given by the kinematic equation:

\[
d_{\text{motorcycle2}} = v_0 \cdot \Delta t + \frac{1}{2} a \cdot (\Delta t)^2
\]

Thus, the total distance covered by the motorcycle after the moment it starts accelerating is:

\[
d_{\text{motorcycle}} = d_{\text{motorcycle1}} + d_{\text{motorcycle2}}
\]

### Step 3: Setting up the equation
To catch up with the car, the total distance covered by the motorcycle must equal the total distance covered by the car plus the initial gap \( d_0 \). Therefore, we set the equation:

\[
v_0 \cdot \Delta t + \frac{1}{2} a \cdot (\Delta t)^2 = d_0
\]

Now we can solve for \( \Delta t \).The time it takes for the motorcycle to catch up with the car after it starts accelerating is approximately \( 3.14 \, \text{seconds} \), rounded to three significant figures.

Would you like further details or have any additional questions?

Here are some related questions:
1. How would the result change if the motorcycle's acceleration was lower?
2. What if the car was also accelerating? How would you model that?
3. How far does the motorcycle travel in total by the time it catches up?
4. How does the initial gap between the vehicles affect the time to catch up?
5. Could the motorcycle catch up if the car was moving faster initially?

**Tip:** When solving motion problems, always start by listing known quantities and the relevant equations for uniform acceleration.

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 18.0 m/s, and the distance between them is 91.0 m. After t1 = 4.00 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. Express the time numerically in seconds using three significant figures.

Solve a kinematics problem where a motorcycle accelerates to catch up with a car traveling at constant speed. Learn how to apply uniform acceleration equations to find the time it takes for the motorcycle to catch up, given an initial speed of 18.0 m/s, a 91.0 m distance, and an acceleration of 7.00 m/s². Suitable for high school physics students in Grades 11-12.

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