# s2a11 - Physics E-1a Homework Solutions#2 Fall 2011 1 From...

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Physics E-1a Homework Solutions #2 Fall 2011 1. From the displacement curve we find the velocities. The maximum value of the slope in the first 5 seconds is about 22.5m 5s = 4.5m / s . The slope between 15 and 25 seconds is about 30m 11s = 2.7m/s . From the velocity curve we find the acceleration. The slope at about 5 seconds is ! 8m / s .2s = ! 40 m s 2 , at 10 seconds 8m / s .2s = 2 and at 15 seconds less than 2 .

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2. This problem can be solved graphically, in which case you would draw a sketch to scale using a protractor and a ruler to find the answer. Here it will be solve analytically by adding components of the vectors. Let the E-W direction be the x-axis and the N-S direction be the y-axis. A X = (12)sin20 ° = 4.1m in neg x ! direction A y = (12)cos20 ° = 11.3m in pos x ! direction B X = (12) cos20 ° = 15.3m in neg x ! direction B y = (12)sin20 ° = 12.9m in neg y ! direction ! R = ! A + ! B ! R x = ( " A x ) + ( " B x ) = " 19.4m and R y = ( + A y ) + ( ! B y ) = ! 1.6m R = R x 2 + R y 2 = 376m 2 + 2.6m 2 = 19.5m tan ! = R y /R x = ( " 1.6m) /( " 19.4m) = 0.082 # ! = 4.7 ° Therefore, you are 19.5 m from the origin in a direction 4.7 ° south of west 3. This is a vector addition problem. Ask the question this way, what velocity do I add to the velocity of the ship (relative to the water v s,w ) to find the velocity of the ship (relative to the earth v s,e )? The answer is the velocity of the water (relative to the earth v w,e ). As a vector addition equation we have , ! v s,w + ! v w ,e = ! v s, e or ! v w,e = ! v s,e ! ! v s,w With v s,w = 4.00m / s 25 ° west of north and v s,e = 4.80m / s 5 ° west of north find v w, e We need the components of the velocities: The N-S component of v s,w = (4.00m / s)cos25 ° = 3.63m / s v s,e = (4.80m / s)cos5 ° = 4.78m / s The E-W component of (note, these components are in the negative x direction and therefore have negative values) N S E W ! = 20 0 ! =40 0 A=12m B=20m A B R
v s,w = (4.00m / s)sin 25 ° = ! 1.69m / s v s,e = (4.80m / s)sin 5

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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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s2a11 - Physics E-1a Homework Solutions#2 Fall 2011 1 From...

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