Calculus 1, Velocity and acceleration

sm121_vel.mws,wdj,7-99

> restart;with(plots):

>

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 constant acceleration, derive the equations of motion. Find the horizonal distance after 1 min 40 sec.

> mach2:=1330*5280/3600;#in feet per second
evalf(mach2);

> given:=diff(x(t),t,t)=a1,diff(y(t),t,t)=a2;#constant acceleration

What functions x(t) and y(t) have the property that their second derivatives are the above constants? To find out, we next integrate the right hand side of the given equations twice.

> v1:=unapply(int(rhs(given[1]),t),t);
v2:=unapply(int(rhs(given[2]),t),t);

> x:=unapply(int(v1(t),t),t);
y:=unapply(int(v2(t),t),t);

> know1:=x(100)=32000;#range at 1min 40sec

> know2:=sqrt(v1(100)^2+v2(100)^2)=mach2;
#speed at 1min 40sec

> a11:=solve(know1,a1);#solve for a1, call it a11

> x0:=(5852/3)/100;
s:=solve(subs(a1=a11,know2),a2);#solve for a2, call it a22

> a22:=s[1];#s[1] or s[2], depending on which is >0
evalf(a22);

> eqns_of_motion:=subs({a1=a11,a2=a22},{x(t),y(t)});

> plot([eqns_of_motion[1],eqns_of_motion[2],t=0..100],scaling=constrained);

A 105 mm Howitzer type M102 cannon used by the US Army firing an HE(M1) projectile has muzzle velocity 494 m/s. The claimed maximum range is 11500 m. Ignoring air resistence, find the predicted maximum range.

> g:=9.8;
x:=(t,s0,theta)->s0*cos(theta)*t;
y:=(t,s0,theta)->s0*sin(theta)*t-(1/2)*g*t^2;

> a1:=plot([x(t,494,Pi/6),y(t,494,Pi/6),t=0..50],color=red):
a2:=plot([x(t,494,Pi/4),y(t,494,Pi/4),t=0..75],color=blue):
a3:=plot([x(t,494,Pi/3),y(t,494,Pi/3.5),t=0..80],color=green):
display(a1,a2,a3);

> t0:=solve(y(t,494,theta)=0,t);

> t_impact:=t0[2];

> range_cannon:=x(t_impact,494,theta);

> Drange:=diff(range_cannon,theta);
theta0:=solve(Drange,theta);

> range0:=subs(theta=theta0[1],x(t_impact,494,theta));
evalf(range0);

>