How to Calculate Maximum Speed, Distance Covered, and Total Time for a Spaceship Journey to the Moon?

[courtrigrad]

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

 

[Leong]

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

 

[quicknote]

for your question. Let's break down each part of the problem and go through the steps to find the answers.

(a) To find the maximum speed attained, we can use the equation v = v0 + at, where v0 is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity is 0 m/s, the acceleration is 15.0 m/s^2, and the time is 600 seconds. Plugging these values into the equation, we get v = 0 + 15.0(600) = 9000 m/s. So, the maximum speed attained is 9000 m/s.

(b) To find the fraction of the total distance traveled at constant speed, we need to first find the distance traveled during the first and last 10 minutes of the trip. During the first 10 minutes, the acceleration is 15.0 m/s^2, so we can use the equation x = x0 + v0t + 1/2at^2 to find the distance traveled. Since x0 and v0 are both 0, we can simplify the equation to x = 1/2at^2. Plugging in the values, we get x = 1/2(15)(600)^2 = 2.7 \times 10^7 m. Similarly, during the last 10 minutes, the acceleration is -15.0 m/s^2, so the distance traveled is also 2.7 \times 10^7 m. Therefore, the total distance traveled at constant speed is 400,000 km - 2(2.7 \times 10^7 m) = 3.9 \times 10^8 m. To find the fraction, we divide this distance by the total distance traveled, which gives us 3.9 \times 10^8 m / 4.0 \times 10^8 m = 0.975, or 97.5% of the total distance.

(c) To find the total time required for the trip, we can add the times for the two separate accelerations. The first acceleration takes 10 minutes, or 600 seconds, and the last acceleration also takes 10 minutes, so the total time is 1200 seconds. However, we also need to account for the time spent traveling at constant speed. Since we know that the spaceship travels