courtrigrad
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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 [itex]4.0 \times 10^{8} m[/itex]. I also converted time to seconds ot [itex]t = 600 s[/itex]. So we know [itex]x_{0} = 0, v_{x}_{0} = 0, x = 4.0 \times 10^{8} m, a_{x} = 15.0 m/sec^{2}, t = 600 sec[/itex]. So would the maximum speed be [itex]v_{x} = 0 + 15(600) = 9000 m/s[/itex]? 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
So first think I did was to convert 400,000 km to meters to get [itex]4.0 \times 10^{8} m[/itex]. I also converted time to seconds ot [itex]t = 600 s[/itex]. So we know [itex]x_{0} = 0, v_{x}_{0} = 0, x = 4.0 \times 10^{8} m, a_{x} = 15.0 m/sec^{2}, t = 600 sec[/itex]. So would the maximum speed be [itex]v_{x} = 0 + 15(600) = 9000 m/s[/itex]? 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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