+21 votes
in Classical Mechanics by
edited by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
+7 votes
by

Constant 1g acceleration refers to the acceleration experienced by an object moving under the influence of a constant force equal to the acceleration due to gravity on Earth's surface, which is approximately 9.8 meters per second squared (m/s^2).

In the scenario you provided, where the rocket is capable of traveling 22 mph in the first minute, 44 mph in the next minute, and 66 mph in the subsequent minute, it means that the rocket is accelerating at a rate equivalent to 1g, or approximately 9.8 m/s^2, throughout this time period.

To understand how the rocket achieves these velocities, we can calculate the distance covered during each minute. Using the constant acceleration equation:

d = v0*t + (1/2)at^2

Where: d = distance covered v0 = initial velocity a = acceleration t = time

Let's calculate the distance covered in each minute:

For the first minute: v0 = 0 mph (starting from rest) t = 1 minute = 60 seconds a = 9.8 m/s^2 ≈ 32.2 ft/s^2 (approximate conversion from meters to feet) Converting mph to ft/s: 22 mph ≈ 32.26 ft/s

Plugging in the values: d = 0t + (1/2)32.260^2 = 0 + 0.532.2*3600 ≈ 349,920 feet ≈ 105,208 meters

Therefore, the rocket covers approximately 349,920 feet (or 105,208 meters) in the first minute.

For the second minute: v0 = 22 mph (starting velocity from the first minute) t = 1 minute = 60 seconds a = 9.8 m/s^2 ≈ 32.2 ft/s^2 (same as before)

Plugging in the values: d = 2260 + 0.532.260^2 = 1,320 + 0.532.2*3600 ≈ 1,319,520 feet ≈ 399,303 meters

Therefore, the rocket covers approximately 1,319,520 feet (or 399,303 meters) in the second minute.

For the third minute: v0 = 44 mph (starting velocity from the second minute) t = 1 minute = 60 seconds a = 9.8 m/s^2 ≈ 32.2 ft/s^2 (same as before)

Plugging in the values: d = 4460 + 0.532.260^2 = 2,640 + 0.532.2*3600 ≈ 2,789,520 feet ≈ 849,399 meters

Therefore, the rocket covers approximately 2,789,520 feet (or 849,399 meters) in the third minute.

Overall, with a constant 1g acceleration, the rocket covers increasingly larger distances in each subsequent minute due to the cumulative effect of acceleration.

Welcome to Physicsgurus Q&A, where you can ask questions and receive answers from other members of the community.
...