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

This preview shows pages 1–2. Sign up to view the full content.

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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## 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.

### Page1 / 4

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online