Question

A glider on an air track moves in the +x direction with a constant acceleration. It has two flags, each exactly 25.4 mm long, with the midpoints of the flags separated by 162 mm. The first flag interrupts the photogate timer for a time 50 ms, and the second flag interrupts the photogate timer for a time 45 ms. ------------- 1. What was the average velocity of the glider during the interval when the first flag was interrupted?------- 2. What was the average velocity of the glider during the interval when the second flag was interrupted?

Answer 1

1 The average velocity for first interruption is
`u = 0.508m/s`

2 The average velocity for second interruption is
`v = 0.564 m/s`

Step-by-step explanation:

From the question we are told that

The length of each the flags is
`L = 25.4 mm = (25.4)/(1000)=0.0254m`

The distance of separation at the mid-point
`d = 162 mm = (162)/(1000) = 0.162 m`

The interruption time for the first flag is
`t_1 = 50ms = 50*10^(-3) s`

The interruption time for the second flag is
`t_2 = 45ms = 45 *10^(-3)s`

The Initial velocity for this motion is obtained mathematically as

`Initial \ velocity (u) = (L)/(t_1)`

L is the distance traveled because the photogate timer measure time at that particular distance

Substituting value

`u = \frac{0.0254}{50*10{-3}}`

`u = 0.508m/s`

The final velocity is mathematically evaluated as

`v = (L)/(t_2)`

Substituting value

`v = \frac{0.0254}{45*10^ {-3}}`

`v = 0.564 m/s`