Answer:
1 The average velocity for first interruption is [tex]u = 0.508m/s[/tex]
2 The average velocity for second interruption is [tex]v = 0.564 m/s[/tex]
Explanation:
From the question we are told that
The length of each the flags is [tex]L = 25.4 mm = \frac{25.4}{1000}=0.0254m[/tex]
The distance of separation at the mid-point [tex]d = 162 mm = \frac{162}{1000} = 0.162 m[/tex]
The interruption time for the first flag is [tex]t_1 = 50ms = 50*10^{-3} s[/tex]
The interruption time for the second flag is [tex]t_2 = 45ms = 45 *10^{-3}s[/tex]
The Initial velocity for this motion is obtained mathematically as
[tex]Initial \ velocity (u) = \frac{L}{t_1}[/tex]
L is the distance traveled because the photogate timer measure time at that particular distance
Substituting value
[tex]u = \frac{0.0254}{50*10{-3}}[/tex]
[tex]u = 0.508m/s[/tex]
The final velocity is mathematically evaluated as
[tex]v = \frac{L}{t_2}[/tex]
Substituting value
[tex]v = \frac{0.0254}{45*10^ {-3}}[/tex]
[tex]v = 0.564 m/s[/tex]
