07_freefall

# Vinitial was 20 ms after 2 sec a was 10 ms as always

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Unformatted text preview: 10 m/s2 1 s = 30 m/s 01s 12s 23s 34s 45s 10 10 10 10 10 10 20 30 40 50 Spring 2008 18 UCSD Physics 10 Starting from rest, letting go: Time Interval Acceleration (m/s2 down) InitFinal Velocity (m/s down) Average Velocity (m/s down) The average velocity in the interval is just Vavg = (vinitial + vfinal) 01s 12s 23s 34s 45s 10 10 10 10 10 0 10 10 20 20 30 30 40 40 50 5 For the 1 2 s interval, 15 25 35 45 vi = 10 m/s vf = 20 m/s So vavg = (10+20) m/s = 15 m/s Spring 2008 19 UCSD Physics 10 Starting from rest, letting go: Time Interva l 01s 12s 23s 34s 45s Acceleration Final (m/s2 down) Velocity (m/s down) 10 10 10 10 10 10 20 30 40 50 Average Velocity (m/s down) 5 15 25 35 45 Dist. moved (m down) 5 15 25 35 45 + Final Position (m down) 5 20 45 80 125 Spring 2008 20 UCSD Physics 10 This can all be done in shorthand The velocity at the end of an interval is just (the starting velocity) plus (the time interval times the acceleration): vfinal = v(t) = vinit + at The position at the end of an interval is just (the starting position) plus (the time interval times the average velocity over the interval): x(t) = xinit + vavgt Since vavg = (vinit + vfinal), and vfinal = vinit + at, = xinit + (vinit + vinit + at) t, or x(t) = xinit + vinitt + at2 Spring 2008 21 x(t) UCSD Physics 10 An aside on units and cancellation What happens if you multiply an acceleration by a time? Units of acceleration are m/s2, units of time are s m Result is m s = s s s = m s s2 And this has units of veloci...
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