A world's land speed record was set by Colonel John P. Stapp when in March 1954 he rode a rocket-propelled sled that moved along a track at 1020 km/h. He and the sled were brought to a stop in 1.4s. ( See Fig. 2-7.) In terms of g, what acceleration did he experience while stopping ?

Answer

To solve the problem, we need to determine the acceleration experienced by Colonel John P. Stapp while stopping the rocket-propelled sled, expressed in terms of \( g \) (acceleration due to gravity).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial speed \( U = 1020 \) km/h
- Final speed \( V = 0 \) (since the sled comes to a stop)
- Time taken to stop \( t = 1.4 \) s

2. **Convert the Initial Speed from km/h to m/s:**
- To convert km/h to m/s, we use the conversion factor:
\[ 1 \text{ km/h} = \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{1}{3.6} \text{ m/s} \]
- Therefore,
\[ U = 1020 \text{ km/h} = 1020 \times \frac{1}{3.6} \text{ m/s} \approx 283.33 \text{ m/s} \]

3. **Calculate the Acceleration:**
- The formula for acceleration \( a \) is given by:
\[ a = \frac{V - U}{t} \]
- Substituting the values:
\[ a = \frac{0 - 283.33}{1.4} \approx \frac{-283.33}{1.4} \approx -202.38 \text{ m/s}^2 \]
- The negative sign indicates deceleration (retardation).

4. **Express the Acceleration in Terms of \( g \):**
- The standard acceleration due to gravity \( g \approx 9.8 \text{ m/s}^2 \).
- To express \( a \) in terms of \( g \):
\[ a = \frac{-202.38}{9.8} g \approx -20.65 g \]
- Rounding this, we can say:
\[ a \approx -21 g \]

5. **Final Answer:**
- The acceleration experienced by Colonel John P. Stapp while stopping is approximately \( 21g \).