Solve Physics Problems: Moon Feather Drop, Bike Acceleration & Horse Gallop

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SUMMARY

This discussion focuses on solving three physics problems involving motion: the drop of a feather on the moon, the acceleration of a bike, and the movement of a horse. For the feather, using the formula d = v_0t + (at²)/2, it takes approximately 0.83 seconds to hit the moon's surface from a height of 1.1 meters, given the moon's gravity of 1.62 m/s². The bike accelerates to 4.5 m/s in 4.5 seconds, covering a total distance of 40.5 meters. The horse canters 150 meters in 18 seconds and gallops back halfway in 4.3 seconds, demonstrating constant acceleration principles.

PREREQUISITES
  • Understanding of kinematics and equations of motion
  • Familiarity with the concept of acceleration
  • Knowledge of gravitational effects on different celestial bodies
  • Ability to perform basic algebraic calculations
NEXT STEPS
  • Study the equations of motion in detail, particularly the SUVAT equations
  • Learn how to calculate acceleration and distance in various motion scenarios
  • Explore the effects of gravity on different planets and celestial bodies
  • Practice solving real-world physics problems involving motion and acceleration
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding motion dynamics and solving related problems in kinematics.

4everYOURS
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Ok there are three of us trying to figure out these problems and we just can't seem to figure out any of them so it'd be the greatest thing in the world if some one could enlighten us on what we're doing wrong. Here are the problems:

13. An astronaut drops a feather from 1.1 m above the surface of the moon. If the acceleration of gravity on the moon is 1.62 m/s2 downward, how long does it take the feather to hit the moon's surface?

2. A bike first accelerates from 0.0 m/s to 4.5 m/s in 4.5 s, then continues at this constant speed for another 3.5 s. What is the total distance traveled by the bike?

5. A horse canters away from its trainer in a straight line, moving 150 m away in 18.0 s. It then turns abruptly and gallops halfway back in 4.3 s.

Help would be sooo much appreciated.
 
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13. use the suvat eqn of motion for distance traveled under a constant acceleration.

2. work out the acceleration during the first stage of movement, hence distance travelled.

5. there's no question.
 
constant acceleration.

You need to use the following formula:
[tex]d = v_0t + \frac{at^2}{2}[/tex]
 

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