Higher 1A T2 2012 Solutions

# 75 s1 substitute into eqn2 eqn2 1275 s1

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Unformatted text preview: p;!:(!&amp;8+!81B(0(23&amp;'.!0+23!#0(1,\$23! &amp;+!/+2.&amp;*I/&amp;\$B(!\$2&amp;(*-(*(2/(%! S2P &quot; S1P = 13 &quot; 12 = 1m thus # = 1/2 = 0.5 m !!!!!!!!!!!!!! v = #f !! v 350 f= = = 700 Hz # 0.5 !!!!!!!!!!#\$\$%!K+*!&quot;&gt;E!)(*\$+,.!,\$--(*(2/(J!&amp;'(!2(8!)1&amp;'!,\$--(*(2/(!5I.&amp;!:(!&quot;&gt;E! 81B(0(23&amp;'.!0+23!!!!#8'\$/'!,+(.!0(1,!&amp;+!,(.&amp;*I/&amp;\$B(!\$2&amp;(*-(*(2/(%&gt;!L+&amp;'!!,\$.&amp;12/(.! ! C&quot;M!12,!CGM!8\$00!\$2/*(1.(!:4!.&quot;!12,!.G&gt;! (S2P + s2) &quot; (S1P + s1) = 0.75 from Pythagoras : eqn(1) (S2P + s2) 2 = 25 + (S1P + s1) 2 (S2P + s2) = 12.75 + s1 !!!!!!! Substitute into eqn(2) eqn(2) ! (12.75 + s1) 2 = (12 + s1) 2 + 25 which can be solved for s1 The answer is then 12 + s1 First find s1: &quot;3.2 s1 = = 4.29 # 4.3 m &quot;0.75 !!!!!!!!!!!!!! new distance from S2 : 13 + s2 = 12.75 + 4.3! ! new distance from S2 = 17 m (c) Have to do (ii) before (i) to find the train speed. ! (ii) If the frequency appears lower to the observer, the source and the observer must be moving away from each other: their velocities will be negative # v...
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