A glider of length 12.3 cm moves on an air track with constant acceleration. A time interval of 0.637 s elapses between the moment when its front end passes a fixed point A along the track and the moment when its back end passes this point. Next, a time interval of 1.27 s elapses between the moment when the back end of the glider passes point A and the moment when the front end of the glider passes a second point B farther down the track. After that, an additional 0.354 s elapses until the back end of the glider passes point B.

(b) Find the acceleration of the glider.

acc) 1.587

frm ntu?

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

First, let's identify the relevant quantities:
- Length of the glider (l) = 12.3 cm = 0.123 m
- Time interval between the front end passing point A and the back end passing A (t₁) = 0.637 s
- Time interval between the back end passing point A and the front end passing point B (t₂) = 1.27 s
- Time interval between the back end passing point B and the front end passing point B (t₃) = 0.354 s

Now, we can use the equations of motion to find the acceleration.

For the first time interval (t₁), we can use the equation:
l = (1/2) * a * t₁²

Rearranging the equation, we get:
a = (2 * l) / (t₁²)

Substituting the values, we have:
a = (2 * 0.123 m) / (0.637 s)²

Next, for the second time interval (t₂), we can use the equation:
l = v₁ * t₂ + (1/2) * a * t₂²

Since the glider is still moving with the same acceleration, v₁ can be considered the velocity at the end of the first interval, which can be calculated using:
v₁ = a * t₁

Substituting the values, we have:
v₁ = a * 0.637 s
Then, we substitute v₁ and the other values into the equation for t₂:
0.123 m = (a * 0.637 s) * 1.27 s + (1/2) * a * (1.27 s)²

Finally, for the third time interval (t₃), we can again use the equation:
l = v₂ * t₃ + (1/2) * a * t₃²

Since the glider is still moving at the same acceleration, v₂ can be considered the velocity at the end of the second interval, which can be calculated using:
v₂ = v₁ + a * t₂

Substituting the values, we have:
v₂ = (a * 0.637 s) + a * 1.27 s
Then, we substitute v₂ and the other values into the equation for t₃:
0.123 m = [(a * 0.637 s) + a * 1.27 s] * 0.354 s + (1/2) * a * (0.354 s)²

We have three equations with the variable 'a', and we can solve them simultaneously to find the value of acceleration.