A glider of length 10.1 cm moves on an air track with constant acceleration. A time interval of 0.732 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 1.01 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 0.336 s elapses until the back end of the glider passes point circled B.

(b) Find the acceleration of the glider.

(c) Explain how you can compute the acceleration without knowing the distance between points circled A and circled B.

To find the acceleration of the glider, we can use the equations of motion.

(b) The time interval, t1, between the moment when the front end passes point circled A and the moment when the back end passes point A is given as 0.732 s.

Using the equation,

d = v₀t + 0.5at²

where d is the distance traveled, v₀ is the initial velocity, t is the time interval, a is the acceleration,

The distance traveled by the glider in this time interval is equal to its length, 10.1 cm. Since the glider moves with constant acceleration, its initial velocity, v₀, is zero.

So we can rewrite the equation as:

10.1 = 0.5at₁²

Solving for acceleration, a:

a = 2*d/t₁²

Plugging in the values, we get:

a = 2 * 10.1 cm / (0.732 s)²

Evaluating the expression, we can find the acceleration of the glider.

(c) To compute the acceleration without knowing the distance between points circled A and circled B, we can use the fact that the glider has the same acceleration throughout the motion.

From the given information, we have:

t₁ = 0.732 s (time interval between the moment when the front end passes point circled A and the moment when the back end passes point A)

t₂ = 1.01 s (time interval between the moment when the back end of the glider passes point circled A and the moment when the front end passes point circled B)

t₃ = 0.336 s (time interval between the moment when the back end passes point circled B)

Let's assume that the acceleration of the glider is denoted as a.

Using the equation v = u + at, where v is the final velocity, u is the initial velocity, and t is the time interval, we can find the final velocity of the glider at different time intervals.

1. From the first interval (t₁), we know that the glider starts from rest (u = 0), and the time interval is t₁ = 0.732 s. Thus, the final velocity at the end of t₁, denoted as v₁, can be found using v₁ = u + at:

v₁ = 0 + a * t₁ = a * 0.732 -- (1)

2. From the second interval (t₂), the glider starts from rest at point A, and the final velocity at the end of t₂, denoted as v₂, can be found using v₂ = u + a * t₂:

v₂ = 0 + a * t₂ = a * 1.01 -- (2)

3. From the third interval (t₃), the glider again starts from rest at point B, and the final velocity at the end of t₃, denoted as v₃, can be found using v₃ = u + a * t₃:

v₃ = 0 + a * t₃ = a * 0.336 -- (3)

Since the glider has the same acceleration throughout the motion, the final velocity v₁ at the end of t₁ is equal to the initial velocity at the start of t₂, v₂:

v₁ = v₂ -- (4)

Similarly, the final velocity v₂ at the end of t₂ is equal to the initial velocity at the start of t₃, v₃:

v₂ = v₃ -- (5)

By equating equations (1) and (2), and equations (2) and (3) using equations (4) and (5), we can eliminate the unknown variables v₁, v₂, and v₃, which helps us compute the acceleration, a.