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 = 0.230 s and t2 = 0.517 s?
To find the magnitude of the coin's displacement between t1 = 0.230 s and t2 = 0.517 s, we first need to determine the acceleration of the coin while falling.
We know that the initial velocity is zero since the coin starts from rest. The standard equation to calculate displacement during uniform acceleration is:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity (0 in this case)
a = acceleration of the coin
t = time
To find the acceleration, we can use another equation:
v = u + at
Where:
v = final velocity (which is the velocity of the coin at time t2)
u = initial velocity (0 in this case)
a = acceleration of the coin
t = time (t2 in this case)
We know that the final velocity is essentially the velocity at t2. However, since the coin is falling, we can assume it is subjected to the acceleration due to gravity (g), which is approximately -9.8 m/s^2. Therefore, the equation becomes:
v = 0 - 9.8(t2 - 0) = -9.8t2
Now, we can substitute this value of v into the displacement equation:
s = ut2 + (1/2)at2^2
Since the initial velocity is zero, the equation simplifies to:
s = (1/2)at2^2
Now, substitute the value of a, which is -9.8 m/s^2, and the given time interval t1 = 0.230 s and t2 = 0.517 s:
s = (1/2)(-9.8)(0.517^2 - 0.230^2)
Now, we can calculate the magnitude of the coin's displacement between t1 = 0.230 s and t2 = 0.517 s using this equation.
To find the magnitude of the coin's displacement, we need to determine the distance it has fallen between t1 = 0.230 s and t2 = 0.517 s.
We can use the kinematic equation for displacement:
Δy = v₀t + 0.5at²
Where:
Δy is the displacement
v₀ is the initial velocity (in this case, 0 since the coin starts at rest)
t is the time interval
a is the acceleration (in this case, due to gravity -9.8 m/s²)
Plugging in the values, we have:
Δy = (0)(0.517-0.230) + 0.5(-9.8)(0.517-0.230)
Simplifying the equation:
Δy = 0.5(-9.8)(0.287)
Calculating the value:
Δy = -1.4096 meters
Since displacement is defined as a vector quantity and represents the distance and direction of an object's change in position, we take the magnitude of the displacement to only consider the distance traveled:
Magnitude of displacement = |Δy| = 1.4096 meters
Therefore, the magnitude of the coin's displacement between t1 = 0.230 s and t2 = 0.517 s is 1.4096 meters.