Answer :
To find the distance covered by the spaceship, we need to follow these steps:
1. Determine the acceleration:
We know the initial velocity ([tex]\(v_i\)[/tex]), final velocity ([tex]\(v_f\)[/tex]), and time ([tex]\(t\)[/tex]):
- [tex]\(v_i = 58.0\)[/tex] meters/second
- [tex]\(v_f = 153.0\)[/tex] meters/second
- [tex]\(t = 12.0\)[/tex] seconds
The formula for acceleration ([tex]\(a\)[/tex]) when the velocities and time are known is:
[tex]\[ a = \frac{v_f - v_i}{t} \][/tex]
Substituting the given values:
[tex]\[ a = \frac{153.0 - 58.0}{12.0} \][/tex]
[tex]\[ a \approx 7.916666666666667 \, \text{meters/second}^2 \][/tex]
2. Calculate the distance:
The formula for distance ([tex]\(d\)[/tex]) covered under uniform acceleration is:
[tex]\[ d = v_i \cdot t + \frac{1}{2} a \cdot t^2 \][/tex]
Using the known values and the calculated acceleration:
[tex]\[ d = 58.0 \cdot 12.0 + \frac{1}{2} \cdot 7.916666666666667 \cdot (12.0)^2 \][/tex]
Breaking it into parts:
[tex]\[ \text{First term: } 58.0 \cdot 12.0 = 696.0 \, \text{meters} \][/tex]
[tex]\[ \text{Second term: } \frac{1}{2} \cdot 7.916666666666667 \cdot 144.0 = 570.0 \, \text{meters} \][/tex]
Adding these two parts together:
[tex]\[ d = 696.0 + 570.0 = 1266.0 \, \text{meters} \][/tex]
So, the correct answer is:
[tex]\[ \boxed{1.27 \times 10^3 \, \text{meters}} \][/tex]
Therefore, the correct choice is:
B. [tex]\(1.27 \times 10^3\)[/tex] meters
1. Determine the acceleration:
We know the initial velocity ([tex]\(v_i\)[/tex]), final velocity ([tex]\(v_f\)[/tex]), and time ([tex]\(t\)[/tex]):
- [tex]\(v_i = 58.0\)[/tex] meters/second
- [tex]\(v_f = 153.0\)[/tex] meters/second
- [tex]\(t = 12.0\)[/tex] seconds
The formula for acceleration ([tex]\(a\)[/tex]) when the velocities and time are known is:
[tex]\[ a = \frac{v_f - v_i}{t} \][/tex]
Substituting the given values:
[tex]\[ a = \frac{153.0 - 58.0}{12.0} \][/tex]
[tex]\[ a \approx 7.916666666666667 \, \text{meters/second}^2 \][/tex]
2. Calculate the distance:
The formula for distance ([tex]\(d\)[/tex]) covered under uniform acceleration is:
[tex]\[ d = v_i \cdot t + \frac{1}{2} a \cdot t^2 \][/tex]
Using the known values and the calculated acceleration:
[tex]\[ d = 58.0 \cdot 12.0 + \frac{1}{2} \cdot 7.916666666666667 \cdot (12.0)^2 \][/tex]
Breaking it into parts:
[tex]\[ \text{First term: } 58.0 \cdot 12.0 = 696.0 \, \text{meters} \][/tex]
[tex]\[ \text{Second term: } \frac{1}{2} \cdot 7.916666666666667 \cdot 144.0 = 570.0 \, \text{meters} \][/tex]
Adding these two parts together:
[tex]\[ d = 696.0 + 570.0 = 1266.0 \, \text{meters} \][/tex]
So, the correct answer is:
[tex]\[ \boxed{1.27 \times 10^3 \, \text{meters}} \][/tex]
Therefore, the correct choice is:
B. [tex]\(1.27 \times 10^3\)[/tex] meters
