1. A car accelerates from 12 m/s to 25 m/s in 6.0 s. What was its acceleration? How far did it travel in this time?

2. A car slows down from 20 m/s to rest in a distance of 85 meters. What was its acceleration?

3. Ando traveling 50 m/s passes a stationary police officer. The police officer immediately begins pursuit at a constant acceleration of 5 m/s2.
a) How much time will it take the officer to reach Ando, assuming that Ando maintains a constant speed?
b) How fast will the police officer be traveling at this time?

Let's tackle each problem one by one.

### 1. Car acceleration and distance traveled
**Given:**
- Initial velocity (\(u\)) = 12 m/s
- Final velocity (\(v\)) = 25 m/s
- Time (\(t\)) = 6.0 s

#### Acceleration (\(a\))
The formula to calculate acceleration is:
\[ a = \frac{v - u}{t} \]

Substitute the given values:
\[ a = \frac{25 \, \text{m/s} - 12 \, \text{m/s}}{6 \, \text{s}} \]
\[ a = \frac{13 \, \text{m/s}}{6 \, \text{s}} \]
\[ a = 2.17 \, \text{m/s}^2 \]

#### Distance traveled (\(s\))
The formula to calculate distance when initial and final velocities, and time are known is:
\[ s = ut + \frac{1}{2}at^2 \]

Substitute the known values:
\[ s = (12 \, \text{m/s} \times 6 \, \text{s}) + \frac{1}{2} \times 2.17 \, \text{m/s}^2 \times (6 \, \text{s})^2 \]
\[ s = 72 \, \text{m} + \frac{1}{2} \times 2.17 \times 36 \]
\[ s = 72 \, \text{m} + 39.06 \, \text{m} \]
\[ s = 111.06 \, \text{m} \]

### 2. Car deceleration
**Given:**
- Initial velocity (\(u\)) = 20 m/s
- Final velocity (\(v\)) = 0 m/s (since the car comes to rest)
- Distance (\(s\)) = 85 m

#### Acceleration (\(a\))
The formula to calculate acceleration when initial velocity, final velocity, and distance are known is:
\[ v^2 = u^2 + 2as \]
Rearrange to solve for \(a\):
\[ a = \frac{v^2 - u^2}{2s} \]

Substitute the known values:
\[ a = \frac{0^2 - (20 \, \text{m/s})^2}{2 \times 85 \, \text{m}} \]
\[ a = \frac{-400 \, \text{m}^2/\text{s}^2}{170 \, \text{m}} \]
\[ a = -2.35 \, \text{m/s}^2 \]

### 3. Police pursuit problem

**Given:**
- Ando's speed (\(v_A\)) = 50 m/s
- Police officer's acceleration (\(a_P\)) = 5 m/s\(^2\)

#### (a) Time for the officer to reach Ando
The distance traveled by Ando (\(s_A\)) in time \(t\) is:
\[ s_A = v_A \cdot t \]

The distance traveled by the police officer (\(s_P\)) in time \(t\) is:
\[ s_P = \frac{1}{2}a_P t^2 \]

When the officer reaches Ando, these distances are equal:
\[ s_A = s_P \]
\[ 50t = \frac{1}{2} \times 5 \times t^2 \]
\[ 50t = 2.5t^2 \]
\[ t^2 = \frac{50t}{2.5} \]
\[ t^2 = 20t \]
\[ t = 20 \, \text{s} \]

#### (b) Speed of the police officer at this time
The final velocity (\(v_P\)) of the police officer can be calculated using:
\[ v_P = u + at \]
Since the initial velocity (\(u\)) of the officer is 0,
\[ v_P = 0 + (5 \, \text{m/s}^2 \times 20 \, \text{s}) \]
\[ v_P = 100 \, \text{m/s} \]

### Summary

1. **Acceleration**: \(2.17 \, \text{m/s}^2\)
   **Distance traveled**: \(111.06 \, \text{m}\)
2. **Deceleration**: \(-2.35 \, \text{m/s}^2\)
3. **Time for officer to reach Ando**: \(20 \, \text{s}\)
   **Speed of officer at this time**: \(100 \, \text{m/s}\)

Would you like more details or have any questions about these solutions?

#### Related Questions:
1. What is the formula for calculating acceleration?
2. How do you calculate the distance traveled when acceleration is constant?
3. How does initial velocity affect the distance traveled under constant acceleration?
4. What are the different equations of motion for uniformly accelerated motion?
5. How can you calculate deceleration if you know the distance and initial velocity?
6. What is the relationship between velocity and time under constant acceleration?
7. How do you determine the time it takes for two objects with different accelerations to meet?
8. How do you calculate final velocity under constant acceleration?

**Tip**: Always keep track of units and ensure consistency throughout your calculations to avoid errors.

Learn how to solve kinematics problems involving acceleration, distance traveled, and final speeds. Includes solutions to problems where a car accelerates from 12 m/s to 25 m/s in 6.0 s, decelerates from 20 m/s to rest over 85 meters, and a police officer pursues a speeding vehicle at 50 m/s. Suitable for high school students.

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