To solve this problem, we need to use the kinematic equations of motion. Specifically, we can use the following two kinematic equations to find the distance covered by the rocket:
1. [tex]\( v = u + at \)[/tex]
2. [tex]\( s = ut + \frac{1}{2}at^2 \)[/tex]
Where:
- [tex]\( v \)[/tex] = final velocity (445 m/s)
- [tex]\( u \)[/tex] = initial velocity (0 m/s, since the rocket is initially at rest)
- [tex]\( t \)[/tex] = time (4.50 seconds)
- [tex]\( a \)[/tex] = acceleration (99.0 m/s²)
- [tex]\( s \)[/tex] = distance
First, confirm the final velocity using the first equation:
[tex]\[ v = u + at \][/tex]
Given:
[tex]\[ u = 0 \, \text{m/s} \][/tex]
[tex]\[ a = 99.0 \, \text{m/s}^2 \][/tex]
[tex]\[ t = 4.50 \, \text{s} \][/tex]
[tex]\[ v = 0 + (99.0 \times 4.50) \][/tex]
[tex]\[ v = 445.5 \, \text{m/s} \][/tex]
This matches approximately with the given final velocity (445 m/s), allowing us to use these values confidently.
Now, use the second equation to calculate the distance [tex]\( s \)[/tex]:
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
Given:
[tex]\[ u = 0 \, \text{m/s} \][/tex]
[tex]\[ a = 99.0 \, \text{m/s}^2 \][/tex]
[tex]\[ t = 4.50 \, \text{s} \][/tex]
[tex]\[ s = (0 \times 4.50) + \frac{1}{2} (99.0) (4.50)^2 \][/tex]
[tex]\[ s = 0 + 0.5 \times 99.0 \times 20.25 \][/tex]
[tex]\[ s = 49.5 \times 20.25 \][/tex]
[tex]\[ s = 1001.25 \, \text{meters} \][/tex]
Hence, the distance covered by the rocket is approximately 1000 meters.
Therefore, the correct answer is:
B. 1.00 × 10³ meters
