Final answer:
a stone dropped from a cliff falls 4.9 m after 1.0 second, reaches a speed of 9.8 m/s after 1.0 second, has an acceleration of 9.8 m/s^2 at 1.0 second, falls 19.6 m during the second second, and has an acceleration of 9.8 m/s^2 at 2.0 seconds.
Step-by-step explanation:
a. To determine the distance fallen after 1.0 second, we can use the formula: Distance = (Initial Velocity * Time) + (0.5 * Acceleration * Time^2). Since the stone is dropped from rest, the initial velocity is 0 m/s and the acceleration is the acceleration due to gravity, which is approximately 9.8 m/s^2. Plugging in the values, we get:
Distance = (0 * 1.0) + (0.5 * 9.8 * 1.0^2) = 0 + 4.9 = 4.9 m.
b. To find the stone's speed after 1.0 second, we can use the formula: Speed = Initial Velocity + (Acceleration * Time). Since the initial velocity is 0 m/s and the acceleration is 9.8 m/s^2, we have:
Speed = 0 + (9.8 * 1.0) = 9.8 m/s.
c. The acceleration at 1.0 second is the same as the acceleration due to gravity, which is approximately 9.8 m/s^2.
d. To calculate the distance the stone falls during the second second, we can again use the formula: Distance = (Initial Velocity * Time) + (0.5 * Acceleration * Time^2). Since the stone was dropped from rest, the initial velocity is 0 m/s. We can calculate the distance by plugging in the time as 2.0 seconds and using the acceleration due to gravity:
Distance = (0 * 2.0) + (0.5 * 9.8 * 2.0^2) = 0 + 19.6 = 19.6 m.
e. The acceleration at 2.0 seconds is still the same as the acceleration due to gravity, which is approximately 9.8 m/s^2.