A small coin, initially at rest, begins falling. If the clock starts when the coin begins to fall, what is the magnitude of the coin\'s displacement between t1 = 0.1749 s and t2 = 0.437 s?
To determine the magnitude of the coin's displacement between two given times, we need two pieces of information: the initial velocity of the coin and the acceleration acting on it.
Since the coin is initially at rest, its initial velocity, u, is zero. Therefore, we only need to find the acceleration.
The acceleration (a) of an object in free fall near the surface of the Earth can be approximated as the acceleration due to gravity, which is approximately 9.8 m/s².
Now, we can use the following equation of motion to find the displacement (s) of the coin:
s = ut + (1/2)at²
Plugging in the values we have:
u = 0 m/s (initial velocity)
a = 9.8 m/s² (acceleration due to gravity)
t1 = 0.1749 s (initial time)
t2 = 0.437 s (final time)
For the displacement between t1 and t2, we can calculate it as follows:
s = (0)(t2) + (1/2)(9.8)(t2²) - [(0)(t1) + (1/2)(9.8)(t1²)]
Simplifying the equation:
s = 0 + 4.9(t2²) - 0 - 4.9(t1²)
s = 4.9(t2² - t1²)
Substituting the given values:
s = 4.9((0.437)² - (0.1749)²)
Now, we can calculate the magnitude of the displacement by taking the absolute value of s.