A spaceship travels 1220 km with a constant uniform acceleration. How much time is required for the acceleration if the spaceship increases its speed from 11.1 km/s to 11.7 km/s ?

To find the time required for the acceleration, we can use the formula for linear acceleration:

\[v = u + at\]

Where:
- \(v\) is the final velocity,
- \(u\) is the initial velocity,
- \(a\) is the acceleration, and
- \(t\) is the time.

Here, the initial velocity \(u\) is given as 11.1 km/s, and the final velocity \(v\) is given as 11.7 km/s. We need to find the acceleration \(a\) and the time \(t\).

Step 1: Convert the velocities from km/s to m/s since the SI unit for acceleration is m/s².
Given:
\(u = 11.1 \, \text{km/s}\)
\(v = 11.7 \, \text{km/s}\)

To convert kilometers per second (km/s) to meters per second (m/s), multiply by 1000:
\(u = 11.1 \, \text{km/s} \times 1000 = 11100 \, \text{m/s}\)
\(v = 11.7 \, \text{km/s} \times 1000 = 11700 \, \text{m/s}\)

Step 2: Plug in the known values into the formula and solve for the acceleration (\(a\)).

\(v = u + at\) \(\implies 11700 = 11100 + a \cdot t\) \(\implies 600 = a \cdot t\)

Step 3: The spacecraft travels 1220 km. We can use the formula for distance travelled with constant acceleration:

\(s = ut + \frac{1}{2} a t^2\)

Substituting the values:
\(1220000 = 11100 \cdot t + \frac{1}{2} a \cdot t^2\)

Step 4: Rearrange the equation to isolate \(a \cdot t^2\):

\(2440000 = 22200 \cdot t + a \cdot t^2\)

Step 5: We have two equations:
\(600 = a \cdot t\) (from Step 2)
\(2440000 = 22200 \cdot t + a \cdot t^2\)

Step 6: Substitute the value of \(a \cdot t\) from Step 2 into the second equation:

\(2440000 = 22200 \cdot t + 600 \cdot t\)

Step 7: Combine the terms with \(t\):

\(2440000 = (22200 + 600) \cdot t\)

Step 8: Solve for \(t\):

\(t = \frac{2440000}{22800}\)

Step 9: Calculate the value of \(t\):

\(t = 107.02 \, \text{s}\) (approximately)

Therefore, the time required for the acceleration is approximately 107.02 seconds.

There is insufficient information here. The acceleration could be either fast or slow during that interval.