A glider on an air track movies in the +x direction with a constant acceleration

Question

A glider on an air track movies in the +x direction with a constant acceleration. It has two flags, each exactly 25.4 mm lond, with the midpoints of the flags separated by 162 mm. The first flag interrupts the photogate timer for a times for a time 50 ms, and the second flag interrupts the photogate timer for a time 45 ms. What was a average velocity of the glider during the interval when the first flag was interrupted? What was the average velocity of the interval when the second flag was interrupted? Approximately how much time elapsed between the passage of the first flag and the passage of the second flag? Calculate the approximate acceleration of the glider

Explanation / Answer

Assuming that the time it takes to interrupt a flag to get an approximate value of instantaneous velocity at that time;
Initial velocity = V_0 = 25.4/50 = .508 mm/ms
final velocity = V_f = 25.4/45 = .564 mm/ms
Between those velocities the cart moves 162 mm. So use kinematic equation to get acceleration is.

a = (V_f^2 - V_0^2)/2s

= (.318 - .258)/(2)(162)

= .000185 mm/ms^2

the elapsed time;

t = (V_f - V_0)/a

= (.564 - .508)/.000185

= 302.7 ms

= 0.3 sec
1) When the first flag was interrupted, the average velocity = (0+v_0)/2 = 0.508/2 = 0.254 mm/ms.
2)when the second flag was interrupted, the average velocity = (0.508+0.564)/2 = 0.536 mm/ms
3) Time elapsed = 0.3 sec.
4) acceleration = 0.185 m/s^2.

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