Answer :
To find which combination of rocket bodies and engines will result in an acceleration of [tex]\(40 \, \text{m/s}^2\)[/tex] at the start of the launch, we need to use Newton's second law of motion, which states that:
[tex]\[ a = \frac{F}{m} \][/tex]
where [tex]\( a \)[/tex] is acceleration, [tex]\( F \)[/tex] is the force, and [tex]\( m \)[/tex] is the mass. We will calculate the acceleration for each combination of body and engine and check which one meets the required acceleration of [tex]\( 40 \, \text{m/s}^2 \)[/tex].
Step-by-Step Solution:
1. Computation for Body 1 and Engine 1:
- Mass of Body 1, [tex]\( m_1 = 0.500 \, \text{kg} \)[/tex]
- Force of Engine 1, [tex]\( F_1 = 25 \, \text{N} \)[/tex]
- Acceleration:
[tex]\[ a = \frac{F_1}{m_1} = \frac{25}{0.500} = 50 \, \text{m/s}^2 \][/tex]
- This combination does not meet the required acceleration of [tex]\( 40 \, \text{m/s}^2 \)[/tex].
2. Computation for Body 1 and Engine 2:
- Mass of Body 1, [tex]\( m_1 = 0.500 \, \text{kg} \)[/tex]
- Force of Engine 2, [tex]\( F_2 = 20 \, \text{N} \)[/tex]
- Acceleration:
[tex]\[ a = \frac{F_2}{m_1} = \frac{20}{0.500} = 40 \, \text{m/s}^2 \][/tex]
- This combination meets the required acceleration of [tex]\( 40 \, \text{m/s}^2 \)[/tex].
3. Computation for Body 2 and Engine 2:
- Mass of Body 2, [tex]\( m_2 = 1.5 \, \text{kg} \)[/tex]
- Force of Engine 2, [tex]\( F_2 = 20 \, \text{N} \)[/tex]
- Acceleration:
[tex]\[ a = \frac{F_2}{m_2} = \frac{20}{1.5} \approx 13.33 \, \text{m/s}^2 \][/tex]
- This combination does not meet the required acceleration of [tex]\( 40 \, \text{m/s}^2 \)[/tex].
4. Computation for Body 3 and Engine 1:
- Mass of Body 3, [tex]\( m_3 = 0.750 \, \text{kg} \)[/tex]
- Force of Engine 1, [tex]\( F_1 = 25 \, \text{N} \)[/tex]
- Acceleration:
[tex]\[ a = \frac{F_1}{m_3} = \frac{25}{0.750} \approx 33.33 \, \text{m/s}^2 \][/tex]
- This combination does not meet the required acceleration of [tex]\( 40 \, \text{m/s}^2 \)[/tex].
After evaluating all possible combinations, we find that the combination which results in the required acceleration of [tex]\( 40 \, \text{m/s}^2 \)[/tex] is:
Body 1 + Engine 2
[tex]\[ a = \frac{F}{m} \][/tex]
where [tex]\( a \)[/tex] is acceleration, [tex]\( F \)[/tex] is the force, and [tex]\( m \)[/tex] is the mass. We will calculate the acceleration for each combination of body and engine and check which one meets the required acceleration of [tex]\( 40 \, \text{m/s}^2 \)[/tex].
Step-by-Step Solution:
1. Computation for Body 1 and Engine 1:
- Mass of Body 1, [tex]\( m_1 = 0.500 \, \text{kg} \)[/tex]
- Force of Engine 1, [tex]\( F_1 = 25 \, \text{N} \)[/tex]
- Acceleration:
[tex]\[ a = \frac{F_1}{m_1} = \frac{25}{0.500} = 50 \, \text{m/s}^2 \][/tex]
- This combination does not meet the required acceleration of [tex]\( 40 \, \text{m/s}^2 \)[/tex].
2. Computation for Body 1 and Engine 2:
- Mass of Body 1, [tex]\( m_1 = 0.500 \, \text{kg} \)[/tex]
- Force of Engine 2, [tex]\( F_2 = 20 \, \text{N} \)[/tex]
- Acceleration:
[tex]\[ a = \frac{F_2}{m_1} = \frac{20}{0.500} = 40 \, \text{m/s}^2 \][/tex]
- This combination meets the required acceleration of [tex]\( 40 \, \text{m/s}^2 \)[/tex].
3. Computation for Body 2 and Engine 2:
- Mass of Body 2, [tex]\( m_2 = 1.5 \, \text{kg} \)[/tex]
- Force of Engine 2, [tex]\( F_2 = 20 \, \text{N} \)[/tex]
- Acceleration:
[tex]\[ a = \frac{F_2}{m_2} = \frac{20}{1.5} \approx 13.33 \, \text{m/s}^2 \][/tex]
- This combination does not meet the required acceleration of [tex]\( 40 \, \text{m/s}^2 \)[/tex].
4. Computation for Body 3 and Engine 1:
- Mass of Body 3, [tex]\( m_3 = 0.750 \, \text{kg} \)[/tex]
- Force of Engine 1, [tex]\( F_1 = 25 \, \text{N} \)[/tex]
- Acceleration:
[tex]\[ a = \frac{F_1}{m_3} = \frac{25}{0.750} \approx 33.33 \, \text{m/s}^2 \][/tex]
- This combination does not meet the required acceleration of [tex]\( 40 \, \text{m/s}^2 \)[/tex].
After evaluating all possible combinations, we find that the combination which results in the required acceleration of [tex]\( 40 \, \text{m/s}^2 \)[/tex] is:
Body 1 + Engine 2