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

Answer:

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

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

Explanation:

From the question we are told that

     The length of each  the flags is  $L = 25.4 mm = \frac{25.4}{1000}=0.0254m$

      The distance of separation at the mid-point   $d = 162 mm = \frac{162}{1000} = 0.162 m$

        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

                   $Initial \ velocity (u) = \frac{L}{t_1}$

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

                           $v = \frac{L}{t_2}$

Substituting value  

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

                             $v = 0.564 m/s$