s1a11 - Physics E-1a Homework Solutions #1 Fall 2011 ! x 12...

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Physics E-1a Homework Solutions #1 Fall 2011 1 .(a) Speed = ! x ! t = 12km 18 min = 18min " 60min hr = 40 km hr (b) Velocity a vector equals the displacement divided by the time or v = 10.3km 18min = 10.3km 18min ! 60min hr " 35km hr ina direction 25 0 SE (c) The total trip time is t = 7hrs 30 min = 7.5hrs. The total distance traveled = 2 X 12.0 km = 24 km. The average speed equals the total distance traveled divided by the total trip time or the average speed = 24km 7.5hrs = 3.2km hr = The average velocity equals the total displacement or change in position divided by the total trip time. v = ! position ! time = 0km 7.5hrs = 0 2 . Blood is accelerated from rest to velocity of 30cm/s in 1.80 cm. First find the acceleration of the blood. That is, v 2 = v 0 2 + 2a ! x " a = v 2 # v 0 2 2 ! x = ( ) 2 2 1.8cm ( ) = 2 Now use v = v 0 + at ! t = v a = 2 = 0.12s If we assume a heartbeat rate of 60 beats/min or one beat per second, then it’s not unreasonable that the contraction time would be 0.12s or about of the total beat cycle. 3 . We know the swan must accelerate from rest v 0 = 0 to a final velocity v f = and that she can accelerate at a = 2 (a) Use equation 2-10c to find the distance traveled by the swan to reach v f . v 2 = v 0 2 + 2a ! x " ! x = v 2 2a = ( ) 2 2 ( ) = 51m (b) Use equation 2-10a to find the time. v f = v 0 + at ! t = v a = 0.350 m s = 17.1s .
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This note was uploaded on 10/26/2011 for the course PHYSICS E-1a taught by Professor Wolfgangrueckner during the Fall '11 term at Harvard.

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s1a11 - Physics E-1a Homework Solutions #1 Fall 2011 ! x 12...

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