What Are the Key Calculations for a Spaceship Journey to the Moon?

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The discussion focuses on calculating key metrics for a spaceship journey to the Moon, which is approximately 400,000 km away. The spaceship accelerates at 15.0 m/s² for the first 10 minutes, achieving a maximum speed of 9,000 m/s. Participants discuss how to determine the fraction of the total distance traveled at constant speed by subtracting the distances covered during acceleration phases from the total distance. Additionally, to find the total time required for the journey, one must account for the time spent traveling at constant speed. Overall, the calculations involve converting units and applying kinematic equations effectively.
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A spaceship ferrying workers to the moon takes a straight line path from the Earth to the moon, a distance of about 400,000 km. It accelerates at 15.0 m/s^2 for the first 10 minutes of the trip, then travels at a constant speed until the last 10 minutes when it accelerates at - 15.0 m/s^2, just coming to rest as it reaches the moon. (a) What is the maximum speed attained? (b) What fraction of the total distance is traveled at constant speed? (c) What total time is required for the ship?

So first think I did was to convert 400,000 km to meters to get 4.0 \times 10^{8} m. I also converted time to seconds ot t = 600 s. So we know x_{0} = 0, v_{x}_{0} = 0, x = 4.0 \times 10^{8} m, a_{x} = 15.0 m/sec^{2}, t = 600 sec. So would the maximum speed be v_{x} = 0 + 15(600) = 9000 m/s? For part (b) would I just subtract the distance that it travels in its first and last 10 minutes from 400,000 km and divide by 400,000 km (in meters)? Part (c) I would add the times for the two separate accelerations?

Thanks
 
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(a) is correct.
(b) you have the idea.
(c) you also need to find the time taken when it travels at constant speed.
 
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