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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 5 "     " }{TEXT 256 41 "    Calculus 1, Velocity and acceleration
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "sm121
_vel.mws,wdj,7-99" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "restar
t;with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 268 "A Saab Viggen Thunderbolt combat \+
aircraft is capable of going from stationary on the ground to Mach 2 (
1330 mph) at an altitute of 32000 ft in 1 min 40 sec. Assuming constan
t acceleration, derive the equations of motion. Find the horizonal dis
tance after 1 min 40 sec." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
55 "mach2:=1330*5280/3600;#in feet per second\nevalf(mach2);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%&mach2G#\"%_e\"\"$" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+nmm]>!\"'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 65 "given:=diff(x(t),t,t)=a1,diff(y(t),t,t)=a2;#constant acceleratio
n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&givenG6$/-%%diffG6$-%\"xG6#%\"
tG-%\"$G6$F-\"\"#%#a1G/-F(6$-%\"yGF,F.%#a2G" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 182 "What functions x(t) and y(t) have the property that th
eir second derivatives are the above constants? To find out, we next i
ntegrate the right hand side of the given equations twice." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "v1:=unapply(int(rhs(given[1]),t),t)
;\nv2:=unapply(int(rhs(given[2]),t),t);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%#v1GR6#%\"tG6\"6$%)operatorG%&arrowGF(*&%#a1G\"\"\"9$F.F(F(F(
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#v2GR6#%\"tG6\"6$%)operatorG%&ar
rowGF(*&%#a2G\"\"\"9$F.F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 55 "x:=unapply(int(v1(t),t),t);\ny:=unapply(int(v2(t),t),t);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"xGR6#%\"tG6\"6$%)operatorG%&arrowG
F(,$*&%#a1G\"\"\")9$\"\"#\"\"\"#F/F2F(F(F(" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"yGR6#%\"tG6\"6$%)operatorG%&arrowGF(,$*&%#a2G\"\"\"
)9$\"\"#\"\"\"#F/F2F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
40 "know1:=x(100)=32000;#range at 1min 40sec" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&know1G/,$%#a1G\"%+]\"&+?$" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 60 "know2:=sqrt(v1(100)^2+v2(100)^2)=mach2;\n#speed at
 1min 40sec" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&know2G/,$*$-%%sqrtG6
#,&*$)%#a1G\"\"#\"\"\"\"\"\"*$)%#a2GF/F0F1F0\"$+\"#\"%_e\"\"$" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "a11:=solve(know1,a1);#solve \+
for a1, call it a11" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$a11G#\"#K\"
\"&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "x0:=(5852/3)/100;\ns
:=solve(subs(a1=a11,know2),a2);#solve for a2, call it a22" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%#x0G#\"%j9\"#v" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"sG6$,$*$-%%sqrtG6#\"(p*4>\"\"\"#\"\"\"\"#v,$F'#!\"\"F/" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "a22:=s[1];#s[1] or s[2], dep
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>%$a22G,$*$-%%sqrtG6#\"(p*4>\"\"\"#\"\"\"\"#v" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+u$)oU=!\")" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 50 "eqns_of_motion:=subs(\{a1=a11,a2=a22\},\{x(t),y(t)\});" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%/eqns_of_motionG<$,$*&-%%sqrtG6#\"(p*4>\"
\"\")%\"tG\"\"#F,#\"\"\"\"$]\",$*$F-F,#\"#;\"\"&" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 73 "plot([eqns_of_motion[1],eqns_of_motion[2],t=0
..100],scaling=constrained);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 
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`$*pc`CFdp7$$\"1(4*o@*4'ztFdp$\"1r*Q8+wIc#Fdp7$$\"1g*QY6V%\\xFdp$\"1cA
\"*ef_\"p#Fdp7$$\"1C1X^m)y3)Fdp$\"14v9/P24GFdp7$$\"1/PFJcqc%)Fdp$\"1zz
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F(-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%(SCALINGG6#%,CONSTRAINEDG-%+AXESLA
BELSG6$%!GFe[l-%%VIEWG6$%(DEFAULTGFi[l" 1 2 0 1 0 2 9 1 4 1 1.000000 
45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 208 "A 105 m
m Howitzer type M102 cannon used by the US Army firing an HE(M1) proje
ctile has muzzle velocity 494 m/s. The claimed maximum range is 11500 \+
m. Ignoring air resistence, find the predicted maximum range." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "g:=9.8;\nx:=(t,s0,theta)->s0
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{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gG$\"#)*!\"\"" }}{PARA 11 "" 1 "" 
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> " 0 "" {MPLTEXT 1 0 200 "a1:=plot([x(t,494,Pi/6),y(t,494,Pi/6),t=0..
50],color=red):\na2:=plot([x(t,494,Pi/4),y(t,494,Pi/4),t=0..75],color=
blue):\na3:=plot([x(t,494,Pi/3),y(t,494,Pi/3.5),t=0..80],color=green):
\ndisplay(a1,a2,a3);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 
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{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "t0:=solve(y(t,494,theta)=0,t);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#t0G6$\"\"!,$-%$sinG6#%&thetaG$\"+lK
;35!\"(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "t_impact:=t0[2];
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%)t_impactG,$-%$sinG6#%&thetaG$\"
+lK;35!\"(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "range_cannon:
=x(t_impact,494,theta);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%-range_ca
nnonG,$*&-%$cosG6#%&thetaG\"\"\"-%$sinGF)F+$\"+HlK!)\\!\"&" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "Drange:=diff(range_cannon,theta);\n
theta0:=solve(Drange,theta);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'Dra
ngeG,&*$)-%$sinG6#%&thetaG\"\"#\"\"\"$!+HlK!)\\!\"&*$)-%$cosGF*F,F-$\"
+HlK!)\\F0" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'theta0G6$$\"+M;)R&y!#
5$!+M;)R&yF(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "range0:=sub
s(theta=theta0[1],x(t_impact,494,theta));\nevalf(range0);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%'range0G,$*&-%$cosG6#$\"+M;)R&y!#5\"\"\"-%$si
nGF)F-$\"+HlK!)\\!\"&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+lK;!\\#!
\"&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "5 0 0" 
65 }{VIEWOPTS 1 1 0 1 1 1803 }