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Let's solve these problems one by one.

### Problem 41E:
#### (a) Finding the magnitude of acceleration (deceleration):

We are given:
- Initial velocity \( v_0 = 60 \, \text{mi/h} \)
- Final velocity \( v = 0 \, \text{mi/h} \) (since the car is brought to a stop)
- Distance \( d = 43 \, \text{m} \)
- Constant acceleration \( a \) is assumed.

First, convert the initial velocity to SI units (meters per second):
\[
1 \, \text{mi/h} = 0.44704 \, \text{m/s}
\]
\[
v_0 = 60 \times 0.44704 = 26.8224 \, \text{m/s}
\]

Now, use the kinematic equation to find the acceleration:
\[
v^2 = v_0^2 + 2a d
\]
Since the car is stopping (\(v = 0\)):
\[
0 = (26.8224)^2 + 2a(43)
\]
Solving for \( a \):
\[
0 = 719.228 + 86a
\]
\[
a = \frac{-719.228}{86} = -8.36 \, \text{m/s}^2
\]
Thus, the magnitude of acceleration is \( |a| = 8.36 \, \text{m/s}^2 \).

Now, express the acceleration in terms of \( g \), where \( g = 9.8 \, \text{m/s}^2 \):
\[
a_g = \frac{8.36}{9.8} = 0.85 \, g
\]
So, the deceleration is \( 8.36 \, \text{m/s}^2 \) or \( 0.85 \, g \).

#### (b) Stopping time:
To find the stopping time, we can use the equation:
\[
v = v_0 + at
\]
Solving for \( t \):
\[
0 = 26.8224 + (-8.36)t
\]
\[
t = \frac{26.8224}{8.36} = 3.21 \, \text{seconds}
\]

Next, to find how many reaction times this stopping time corresponds to, we are given a reaction time \( T = 400 \, \text{ms} = 0.4 \, \text{seconds} \). So the number of reaction times is:
\[
\frac{3.21}{0.4} = 8.03
\]
Thus, the stopping time corresponds to approximately 8 reaction times.

### Problem 42E:
#### Finding acceleration experienced by Colonel John P. Stapp:

We are given:
- Initial velocity \( v_0 = 1020 \, \text{km/h} \)
- Final velocity \( v = 0 \, \text{km/h} \) (since the sled comes to a stop)
- Time \( t = 1.4 \, \text{seconds} \)

First, convert the initial velocity to SI units (meters per second):
\[
1 \, \text{km/h} = \frac{1000}{3600} \, \text{m/s} = 0.27778 \, \text{m/s}
\]
\[
v_0 = 1020 \times 0.27778 = 283.3336 \, \text{m/s}
\]

Now, use the equation \( a = \frac{v - v_0}{t} \):
\[
a = \frac{0 - 283.3336}{1.4} = -202.38 \, \text{m/s}^2
\]
Thus, the magnitude of acceleration is \( |a| = 202.38 \, \text{m/s}^2 \).

Now, express this in terms of \( g \):
\[
a_g = \frac{202.38}{9.8} \approx 20.65 \, g
\]
So, Colonel Stapp experienced an acceleration of approximately \( 20.65 \, g \).

---

Do you have any questions or would you like more details on any part of the solutions?

Here are some related questions:
1. What other kinematic equations are useful in similar problems?
2. How does varying reaction time affect stopping distance?
3. Can you calculate stopping distances for other speeds using the same deceleration?
4. What are the typical g-forces experienced in everyday situations (like car braking or roller coasters)?
5. How can we measure reaction time in a real-world scenario?

**Tip:** Always make sure to convert units consistently when working with different measurement systems in physics problems.

A car can be braked to a stop from 60 mi/h in 43 m. (a) What is the magnitude of the acceleration (really a deceleration) in SI units and “g units”? (b) What is the stopping time? If your reaction time for braking is 400 ms, to how many “reaction times” does the stopping time correspond?

Solve a kinematics problem involving a car braking from 60 mi/h to a stop in 43 meters. This step-by-step solution calculates the magnitude of deceleration in SI units and g units, as well as stopping time in terms of reaction times, using core kinematic equations. Suitable for high school or college physics students.

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