Review. You are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As soon as the traffic light turns green, your car speeds up from rest to 47.0 mi/h with constant acceleration 8.00 mi/h/s and thereafter moves with a constant speed of 47.0 mi/h. At the same time, the cyclist speeds up from rest to 21.0 mi/h with constant acceleration 15.00 mi/h/s and thereafter moves with a constant speed of 21.0 mi/h. What is the maximum distance (in ft) by which the bicycle leads your car during this time interval?


To solve this problem, we'll need to calculate the positions of both the car and the bicyclist as functions of time, and then find the point where the difference between their positions is greatest.

### Step 1: Define the key information.

- **Car:**
  - Initial speed = 0 mi/h
  - Maximum speed = 47.0 mi/h
  - Acceleration = 8.00 mi/h/s

- **Bicyclist:**
  - Initial speed = 0 mi/h
  - Maximum speed = 21.0 mi/h
  - Acceleration = 15.00 mi/h/s

### Step 2: Convert velocities to ft/s.

We need to work in consistent units. Since the result must be in feet, let's convert the velocities from miles per hour (mi/h) to feet per second (ft/s).

- 1 mi/h = 1.467 ft/s
- Car's maximum speed = \( 47.0 \, \text{mi/h} \times 1.467 \, \text{ft/s per mi/h} = 68.949 \, \text{ft/s} \)
- Bicyclist's maximum speed = \( 21.0 \, \text{mi/h} \times 1.467 \, \text{ft/s per mi/h} = 30.807 \, \text{ft/s} \)

### Step 3: Time taken to reach maximum speeds.

- Car's acceleration = \( 8.00 \, \text{mi/h/s} \times 1.467 \, \text{ft/s per mi/h} = 11.736 \, \text{ft/s}^2 \)
- Bicyclist's acceleration = \( 15.00 \, \text{mi/h/s} \times 1.467 \, \text{ft/s per mi/h} = 22.005 \, \text{ft/s}^2 \)

Now calculate the time for each to reach their maximum speed:

- For the car:
  \[
  t_{\text{car}} = \frac{\text{maximum speed}}{\text{acceleration}} = \frac{68.949 \, \text{ft/s}}{11.736 \, \text{ft/s}^2} = 5.877 \, \text{seconds}
  \]
  
- For the bicyclist:
  \[
  t_{\text{bike}} = \frac{\text{maximum speed}}{\text{acceleration}} = \frac{30.807 \, \text{ft/s}}{22.005 \, \text{ft/s}^2} = 1.400 \, \text{seconds}
  \]

### Step 4: Calculate the distances traveled during acceleration.

The distance covered during acceleration is given by the equation:
\[
d = \frac{1}{2} a t^2
\]
For the car during acceleration:
\[
d_{\text{car, acc}} = \frac{1}{2} \times 11.736 \, \text{ft/s}^2 \times (5.877 \, \text{seconds})^2 = 203.736 \, \text{ft}
\]

For the bicyclist during acceleration:
\[
d_{\text{bike, acc}} = \frac{1}{2} \times 22.005 \, \text{ft/s}^2 \times (1.400 \, \text{seconds})^2 = 21.566 \, \text{ft}
\]

### Step 5: Calculate the distances traveled after reaching maximum speed.

Now, the bicyclist moves with constant speed after 1.4 seconds. The car reaches constant speed at 5.877 seconds. So for the next 4.477 seconds (5.877 - 1.400), the bicyclist will travel at a constant speed of 30.807 ft/s.

- Distance covered by the bicyclist after acceleration:
\[
d_{\text{bike, const}} = 30.807 \, \text{ft/s} \times 4.477 \, \text{seconds} = 137.961 \, \text{ft}
\]

The car, after reaching 5.877 seconds, will not have traveled with constant speed yet in this time interval.

### Step 6: Determine the maximum lead.

To find when the bicycle leads the most, we calculate the relative distance between their positions. Initially, the bicycle accelerates faster, so we expect it to lead until the car catches up due to higher speed. Since the bicyclist’s total distance (at \(t_{\text{bike}} + t_{\text{car}}\)) adds up to \( d_{\text{bike, final distance}}\

You are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As soon as the traffic light turns green, your car speeds up from rest to 47.0 mi/h with constant acceleration 8.00 mi/h/s and thereafter moves with a constant speed of 47.0 mi/h. At the same time, the cyclist speeds up from rest to 21.0 mi/h with constant acceleration 15.00 mi/h/s and thereafter moves with a constant speed of 21.0 mi/h. What is the maximum distance (in ft) by which the bicycle leads your car during this time interval?

Solve a physics problem where a car and a bicyclist accelerate from rest at different rates, then move at constant speeds. Learn how to calculate the maximum distance the bicycle leads the car using kinematic equations. Ideal for high school students in Grades 11-12.

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