To calculate the space shuttle's deceleration, we use the formula for acceleration (or deceleration, which is just a negative acceleration):

$a = \frac{\Delta v}{\Delta t}$

Where:

• $a$ is the acceleration (or deceleration, in this case).
• $\Delta v$ is the change in velocity.
• $\Delta t$ is the time interval.

Given:

• Initial velocity, $v_i = 1200 \, \text{km/hr}$
• Final velocity, $v_f = 400 \, \text{km/hr}$
• Time interval, $\Delta t = 10 \, \text{seconds}$

Step 1: Convert velocities to meters per second (m/s), since the time is in seconds.

1 kilometer per hour ($\text{km/hr}$) is equal to $\frac{1000 \, \text{m}}{3600 \, \text{s}} = \frac{5}{18} \, \text{m/s}$.

• $v_i = 1200 \, \text{km/hr} \times \frac{5}{18} = 333.33 \, \text{m/s}$
• $v_f = 400 \, \text{km/hr} \times \frac{5}{18} = 111.11 \, \text{m/s}$

Step 2: Find the change in velocity, $\Delta v$:

$\Delta v = v_f - v_i = 111.11 \, \text{m/s} - 333.33 \, \text{m/s} = -222.22 \, \text{m/s}$

(The negative sign indicates a deceleration, meaning a reduction in speed.)

Step 3: Calculate the deceleration using the formula:

$a = \frac{\Delta v}{\Delta t} = \frac{-222.22 \, \text{m/s}}{10 \, \text{s}} = -22.22 \, \text{m/s}^2$

Answer:

The space shuttle's deceleration is $22.22 \, \text{m/s}^2$.

Would you like more details or have any questions?

Here are 5 related questions to expand on this:

1. How would the deceleration change if the time interval was longer?
2. What would be the final velocity if the shuttle decelerated for 15 seconds instead of 10?
3. How does the deceleration compare to the acceleration of a car braking from 100 km/hr to 0 in 5 seconds?
4. What is the effect of deceleration on the shuttle's fuel consumption?
5. How could we calculate the total distance traveled during this deceleration?

Tip: Always ensure that your units are consistent when performing calculations, especially when converting between units like km/hr and m/s.