To solve the problem step by step, we will analyze the motion of the rocket in three phases:
1. **Phase 1: Rocket Ascending with Fuel**
2. **Phase 2: Rocket at the Highest Point**
3. **Phase 3: Rocket Descending**
### Step 1: Analyze Phase 1 (Rocket Ascending with Fuel)
- The rocket is fired vertically upwards with a net acceleration of \(4 \, \text{m/s}^2\) and an initial velocity \(u = 0\).
- The time of ascent is \(t = 5 \, \text{s}\).
Using the equation of motion:
\[
v = u + at
\]
where:
- \(v\) = final velocity at the end of phase 1
- \(u = 0\)
- \(a = 4 \, \text{m/s}^2\)
- \(t = 5 \, \text{s}\)
Substituting the values:
\[
v = 0 + (4 \times 5) = 20 \, \text{m/s}
\]
### Step 2: Analyze Phase 2 (Rocket at the Highest Point)
- After 5 seconds, the fuel is finished, and the rocket decelerates due to gravity \(g = 10 \, \text{m/s}^2\).
- The velocity at the highest point becomes \(0 \, \text{m/s}\).
Using the equation of motion again:
\[
0 = v - gt
\]
where:
- \(v = 20 \, \text{m/s}\) (velocity at the end of phase 1)
- \(g = 10 \, \text{m/s}^2\)
Rearranging gives us:
\[
gt = v \implies t = \frac{v}{g} = \frac{20}{10} = 2 \, \text{s}
\]
### Step 3: Calculate Total Time to Reach the Highest Point
The total time taken to reach the highest point (point B) is:
\[
t_{total} = 5 \, \text{s} + 2 \, \text{s} = 7 \, \text{s}
\]
### Step 4: Analyze Phase 3 (Rocket Descending)
- After reaching the highest point, the rocket accelerates downwards with \(g = 10 \, \text{m/s}^2\) until it reaches the ground.
Using the equation for displacement during free fall:
\[
s = ut + \frac{1}{2}gt^2
\]
where:
- \(u = 0\) (velocity at the highest point)
- \(g = 10 \, \text{m/s}^2\)
- \(t\) is the time taken to fall back to the ground.
Let \(t_{fall}\) be the time taken to fall from the highest point to the ground. The total time to reach the ground is \(10 \, \text{s}\).
Using the equation for the distance fallen:
\[
s = \frac{1}{2}gt^2
\]
Substituting \(g = 10 \, \text{m/s}^2\) and \(t = 3 \, \text{s}\) (from \(7 \, \text{s}\) to \(10 \, \text{s}\)):
\[
s = \frac{1}{2} \times 10 \times (3)^2 = 45 \, \text{m}
\]
### Step 5: Calculate Total Displacement
The total displacement from the ground to the highest point (point A) can be calculated using the area under the velocity-time graph:
\[
\text{Displacement from O to A} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 20 = 50 \, \text{m}
\]
### Step 6: Total Displacement
The total displacement when the rocket returns to the ground is:
\[
\text{Total Displacement} = \text{Displacement from O to A} + \text{Displacement from A to B} = 50 \, \text{m} + 45 \, \text{m} = 95 \, \text{m}
\]
### Step 7: Plotting the Graphs
- **Velocity-Time Graph**:
- From \(0\) to \(5 \, \text{s}\), the graph is a straight line increasing from \(0\) to \(20 \, \text{m/s}\).
- From \(5\) to \(7 \, \text{s}\), the graph is a straight line decreasing from \(20 \, \text{m/s}\) to \(0 \, \text{m/s}\).
- From \(7\) to \(10 \, \text{s}\), the graph is a straight line decreasing from \(0 \, \text{m/s}\) back to \(-30 \, \text{m/s}\) (indicating downward motion).
- **Displacement-Time Graph**:
- The graph will show a parabolic shape from \(0\) to \(5 \, \text{s}\) (as the rocket ascends).
- It will then have a linear increase from \(5\) to \(7 \, \text{s}\) (as the rocket decelerates).
- Finally, it will show a steeper linear increase as the rocket falls back to the ground.
