Answer :
Final answer:
The additional distance the marble falls in the next second on another planet, where it fell 4.00 m in the first second, is found using the formula for distance in uniform acceleration (1/2)gt^2. The total distance after two seconds is 16.00 m, subtract the first 4.00 m to get the additional distance of 12.00 m. So, the correct answer is Option c) 12.0 m.
Explanation:
The question involves the physics concept of uniform acceleration in free fall under gravity. Given that marble is released from rest and falls 4.00 m in the first second on another planet, we need to determine the additional distance it falls in the next second. Under constant acceleration, the distance covered by an object in free fall from rest after t seconds is given by the formula s = (1/2)gt^2, where s is the distance, g is the acceleration due to gravity, and t is the time elapsed.
After the first second, the marble has fallen 4.00 m. This distance corresponds to the first term in the sequence of distances for each second under constant acceleration. So for the second, the marble will fall an additional distance corresponding to the term in the sequence for t=2 seconds minus the distance already fallen:
- Total distance after 2 seconds = (1/2)g(2^2)
- Distance fell in the first 1 second = 4.00 m
- Additional distance in the next 1 second = Total after 2 seconds - Distance already fallen
If we denote the acceleration due to gravity on this planet as g, using the given information, we have 4.00 m = (1/2)g(1^2), which means g = 8.00 m/s^2. Therefore, after two seconds of fall, the total distance covered would be (1/2)*8.00*(2^2) = 16.00 m. To find the additional distance fallen during the second second, we subtract the distance fallen in the first second:
16.00 m (after 2 seconds) - 4.00 m (after 1 second) = 12.00 m (additional distance in the second second)
So, the correct answer is Option c) 12.0 m.
