A glider of length 12.7 cm moves on an air track with constant acceleration. A time interval of 0.563 s elapses between the moment when its front end passes a fixed point circled A along the track and the moment when its back end passes this point. Next, a time interval of 1.11 s elapses between the moment when the back end of the glider passes point circled A and the moment when the front end of the glider passes a second point circled B farther down the track. After that, an additional 0.499 s elapses until the back end of the glider passes point circled B.
To solve this problem, we will use the equations of motion for a uniformly accelerated linear motion.
Let's break down the problem into two parts: motion from point A to point B, and motion from point B to the back end of the glider.
Part 1: Motion from point A to point B
First, let's find the acceleration of the glider from the time interval between its front end passing point A and its back end passing point A.
We are given:
- Length of the glider = 12.7 cm
- Time interval from front end passing point A to back end passing point A = 0.563 s
We can use the equation:
d = v₀t + 0.5at²
where:
d = distance traveled (length of the glider)
v₀ = initial velocity (0 m/s, assuming it starts from rest)
t = time interval (0.563 s)
a = acceleration
Rearranging the equation:
d = 0.5at²
2d = at²
2d/t² = a
Substituting the given values:
2(12.7 cm) / (0.563 s)² = a
25.4 cm / (0.316769 s)² = a
25.4 cm / 0.100380021 s² = a
253.294957415 cm/s² = a
Therefore, the glider's acceleration from point A to point B is approximately 253.29 cm/s².
Part 2: Motion from point B to the back end of the glider
Next, let's find the distance traveled from point B to the back end of the glider using the time interval given.
We are given:
- Time interval from the back end passing point B to the front end passing point B = 0.499 s
Using the equation:
d = v₀t + 0.5at²
where:
d = distance traveled (unknown)
v₀ = initial velocity (unknown)
t = time interval (0.499 s)
a = acceleration (253.29 cm/s², obtained from Part 1)
Since the glider is already moving, we can assume its initial velocity is not zero. Therefore, we have two unknowns: d and v₀.
However, we can use another equation to find the initial velocity, v₀:
v = v₀ + at
where:
v = final velocity (unknown)
Since the time interval is the same as Part 1, and the glider is now moving from rest at point B to the front end passing point B, we can use the equation:
v = v₀ + at
0 = v₀ + (253.29 cm/s²)(0.499 s)
Solving for v₀:
0 = v₀ + 126.4 cm/s
v₀ = -126.4 cm/s (negative sign indicates opposite direction)
Now, let's go back to finding the distance traveled:
d = v₀t + 0.5at²
d = (-126.4 cm/s)(0.499 s) + 0.5(253.29 cm/s²)(0.499 s)²
d = -63.02 cm + 31.69 cm
d = -31.33 cm
Since distance cannot be negative, we take the magnitude of the distance traveled:
|d| = 31.33 cm
Therefore, the distance from point B to the back end of the glider is approximately 31.33 cm.
To summarize:
- The glider's acceleration from point A to point B is approximately 253.29 cm/s².
- The distance from point B to the back end of the glider is approximately 31.33 cm.