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A glider on an air track moves in the direction with a constant acceleration. It has two flags, each exactly 0.02 m long, whith the midpoints of the flags separated by 0.150 m. The first flag interrupts the photogate timer for a time 0.058 s, and the second flag interrups the photogate time for a time 0.047 s.

  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.
  3. Approximately how much time elapsed between the passage of the first flag and the passage of the second flag.
  4. Calculate the approximate acceleration of the glider.

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Rodolfo
RodolfoLv5
25 Jan 2021

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

You are a student in a science class that is using the following apparatus to determine the value of g. Two photogates are used. (Note: You may be familiar with photogates in everyday living. You see them in the doorways of some stores. They are designed to ring a bell when someone interrupts the beam while walking through the door.) One photogate is located at the edge of a table that is 1.00 m above the floor, and the second photogate is located directly below the first, at a height 0.500 m above the floor. You are told to drop a marble through these gates, releasing it from rest a negligible distance above the upper gate. The upper gate starts a timer as the ball passes through its beam. The second photogate stops the timer when the ball passes through its beam. (a) Prove that the experimental magnitude of free-fall acceleration is given by gexp = (2?y)/(?t)2, where ?y is the vertical distance between the photogates and ?t is the fall time. (b) For your setup, what value of ?t would you expect to measure, assuming gexp is the standard value (9.81 m/s2)? (c) During the experiment, a slight error is made. Instead of locating the first photogate even with the top of the table, your not-so-careful lab partner locates it 0.50 cm lower than the top of the table. However, she does manage to properly locate the second photogate at a height of 0.50 m above the floor. However, she releases the marble from the same height that it was released from when the photogate was 1.00 m above the floor. What value of gexp will you and your partner determine? What percentage difference does this represent from the standard value of g?

vf = vi + at
rx = vit + ½ at2
vf2 = vi2 +2arx
rx = ½ (vi + vf)t

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I no this is alot of questions but im having trouble answeringthese! So please help me..

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