Select the correct answer. A spaceship moving with an initial velocity of 58.0 meters/second experiences a uniform acceleration and attains a final velocity of 153meters/second What distance has the spaceship covered after 12.0 seconds? A. 6.96times 10^2 meters B. 1.27times 10^3 meters C. 5.70times 10^2 meters D. 1.26times 10^2 meters E. 6.28times 10^2 meters

To find the distance covered by the spaceship, we can use the equation of motion:<br /><br />\[d = ut + \frac{1}{2}at^2\]<br /><br />where:<br />- \(d\) is the distance covered,<br />- \(u\) is the initial velocity,<br />- \(a\) is the acceleration, and<br />- \(t\) is the time.<br /><br />First, we need to find the acceleration. We can use the equation:<br /><br />\[v = u + at\]<br /><br />where:<br />- \(v\) is the final velocity,<br />- \(u\) is the initial velocity,<br />- \(a\) is the acceleration, and<br />- \(t\) is the time.<br /><br />Rearranging the equation to solve for \(a\), we get:<br /><br />\[a = \frac{v - u}{t}\]<br /><br />Substituting the given values, we have:<br /><br />\[a = \frac{153 - 58}{12} = \frac{95}{12} \approx 7.92 \, m/s^2\]<br /><br />Now, we can use the equation of motion to find the distance covered:<br /><br />\[d = 58 \times 12 + \frac{1}{2} \times 7.92 \times 12^2\]<br /><br />Simplifying the equation, we get:<br /><br />\[d = 696 + \frac{1}{2} \times 7.92 \times 144\]<br /><br />\[d = 696 + 571.68\]<br /><br />\[d = 1267.68\]<br /><br />Therefore, the correct answer is B. \(1.27 \times 10^3\) meters.