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Let's solve each question in turn.

---

### Problem 3:
**Given**:  
- Initial speed, \( u = 0 \) (starting from rest)
- Final speed, \( v = 25.0 \, \text{m/s} \)
- Time, \( t = 10.0 \, \text{s} \)

#### (a) Find the acceleration \( a \):
We use the formula:
\[
v = u + at
\]
Solving for \( a \):
\[
a = \frac{v - u}{t} = \frac{25.0 \, \text{m/s} - 0}{10.0 \, \text{s}} = 2.5 \, \text{m/s}^2
\]

#### (b) Find the distance traveled \( s \):
We use the formula:
\[
s = ut + \frac{1}{2} a t^2
\]
Since \( u = 0 \):
\[
s = \frac{1}{2} \times 2.5 \, \text{m/s}^2 \times (10.0 \, \text{s})^2 = 125 \, \text{m}
\]

---

### Problem 4:
**Given**:
- Distance, \( s = \text{quarter mile} = 0.25 \, \text{mi} \)
- Time, \( t = 6.0 \, \text{s} \)

We assume constant acceleration from rest and find the final speed using the relationship:
\[
s = \frac{1}{2} a t^2
\]
Solving for \( a \):
\[
a = \frac{2s}{t^2} = \frac{2 \times 0.25 \, \text{mi}}{(6.0 \, \text{s})^2} = 0.01389 \, \text{mi/s}^2
\]

Now, using \( v = at \) to find the final speed:
\[
v = 0.01389 \, \text{mi/s}^2 \times 6.0 \, \text{s} = 0.08333 \, \text{mi/s}
\]
Converting to miles per hour:
\[
v = 0.08333 \, \text{mi/s} \times 3600 \, \text{s/h} = 300 \, \text{mi/h}
\]

---

### Problem 5:
**Given**:
- Initial speed, \( u = 12.0 \, \text{m/s} \)
- Deceleration, \( a = -6.20 \, \text{m/s}^2 \)
- Reaction time delay, \( t_{\text{delay}} = 0.510 \, \text{s} \)

#### (a) Distance during the reaction time:
\[
s_{\text{reaction}} = u \times t_{\text{delay}} = 12.0 \, \text{m/s} \times 0.510 \, \text{s} = 6.12 \, \text{m}
\]

#### (b) Stopping distance after applying brakes:
We use:
\[
v^2 = u^2 + 2as
\]
Since \( v = 0 \) (the car stops), we solve for \( s \):
\[
s = \frac{-u^2}{2a} = \frac{-(12.0 \, \text{m/s})^2}{2 \times -6.20 \, \text{m/s}^2} = 11.61 \, \text{m}
\]

#### (c) Total stopping distance:
\[
s_{\text{total}} = s_{\text{reaction}} + s = 6.12 \, \text{m} + 11.61 \, \text{m} = 17.73 \, \text{m}
\]

---

### Summary:
1. **Problem 3**: Acceleration \( a = 2.5 \, \text{m/s}^2 \), Distance \( s = 125 \, \text{m} \)
2. **Problem 4**: Racer's speed \( = 300 \, \text{mi/h} \)
3. **Problem 5**: Total stopping distance \( = 17.73 \, \text{m} \)

Would you like more detailed explanations or have additional questions?

---

**Further Questions:**
1. How would the answers change if the initial speed were not zero?
2. What would happen if the deceleration rate in Problem 5 was higher or lower?
3. How do these calculations apply to real-world vehicle safety and design?
4. Can you determine the time required to stop for Problem 5?
5. What other factors could affect the drag racer’s final speed?

**Tip:** Remember to always check units, especially when converting speeds or distances between different systems.

A train accelerates from rest to a speed of 25.0 m/s in 10.0 s. Assuming the acceleration to be constant over this interval find (a) the magnitude of the acceleration, and (b) the distance traveled during this interval.

A drag racer, starting from rest, travels a quarter mile in 6.0 s. What is the racer’s speed (in mi/h) at the end of the race?

Suppose a car traveling at 12.0 m/s sees a traffic light turn red. After 0.510 s have elapsed, the driver applies the brakes, and the car decelerates at 6.20 m/s². What is the stopping distance of the car, as measured from the point where the driver first notices the red light?

Solve kinematics problems including train acceleration, drag racer speed, and car stopping distance with step-by-step solutions. Learn to calculate acceleration, convert units, and apply equations of motion for constant acceleration. Suitable for high school physics students.

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