To determine the distance covered by the rocket, we will use the kinematic equation for uniformly accelerated motion. The equation is:
[tex]\[ \text{distance} = \text{initial velocity} \times \text{time} + \frac{1}{2} \times \text{acceleration} \times (\text{time})^2 \][/tex]
Given values:
- Initial velocity, [tex]\( u = 0 \)[/tex] meters/second (since the rocket is initially at rest)
- Acceleration, [tex]\( a = 99.0 \)[/tex] meters/second[tex]\(^2\)[/tex]
- Time, [tex]\( t = 4.50 \)[/tex] seconds
Substituting these values into the equation:
[tex]\[ \text{distance} = 0 \times 4.50 + \frac{1}{2} \times 99.0 \times (4.50)^2 \][/tex]
Since the initial velocity term [tex]\( 0 \times 4.50 \)[/tex] is [tex]\( 0 \)[/tex], we simplify the equation to:
[tex]\[ \text{distance} = \frac{1}{2} \times 99.0 \times (4.50)^2 \][/tex]
Calculating:
1. First, calculate the square of the time:
[tex]\[ (4.50)^2 = 20.25 \][/tex]
2. Then multiply by the acceleration:
[tex]\[ 99.0 \times 20.25 = 2004.75 \][/tex]
3. Now, divide by 2:
[tex]\[ \frac{2004.75}{2} = 1002.375 \][/tex]
Thus, the distance covered by the rocket is:
[tex]\[ 1002.375 \text{ meters} \][/tex]
Looking at the options given:
A. [tex]\( 2.50 \times 10^2 \)[/tex] meters
B. [tex]\( 1.00 \times 10^3 \)[/tex] meters
C. [tex]\( 5.05 \times 10^2 \)[/tex] meters
D. [tex]\( 2.00 \times 10^3 \)[/tex] meters
E. [tex]\( 1.00 \times 10^2 \)[/tex] meters
The correct answer is:
[tex]\[ \boxed{1.00 \times 10^3 \text{ meters}} \][/tex]
So, the correct answer is [tex]\( B \)[/tex].
