# Hw3 203 - 00 JD C 8 a ^ o o Q X ^ k) i 0 U 6 <T '? j* o...

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Unformatted text preview: 00 JD C 8 a ^ o o Q X ^ k) i 0 U 6 <T '? j* o o CO S 5 O <* >9 3 00 t' 0 1 41 u QL e m u :*> I i + 41 f 2 CO -d E ca 4- u * 6 ^^^ ^ -i e 3 u ^\> O X 1? 9 1 2 u 3 ~ e CB v 10 < 3 x 01 =il O u ^ c il O (* O O fe n f* In -.;j " oo e x3 C^ O1 Tr u \_x ^ CO 1 I ' 4J O 8 3 4 ^s CO Q ^-* 85 \bs. energy O y*_ qa.-?5Vk-ft /ft s * 9 MO 000 k F\r, so ooos-V \rCfc} ? gfc dx = Qt X CO B-- o fto* - " e -hx^o tat SO P* 1 HP * O The is fM -OO6ST3 = FSolve Page 5 / 6 9.43) function homework943() % Problem 9.43 Solution % Feb 5, 2008 % CONSTANTS P=550 ; % power in lbf*ft/s m= 150; %lbm g-32.2; %ft/s A 2 % INTIAL CONDITIONS vO= 0.001; % initial velocity, zero makes the solution explode tspan =[0 1000]; %time interval of integration error = le-4; % Set error tolerance and use 'event detection' options = odeset('abstor, error, 'reltol', error) ; Ask Matlab to SOLVE odes in function 'rhs' [t v] = ode45(@rhs,tspan, vO, options, P, m, g) %UNPACK the zarray (the solution) into sensible variables plot (t,v) title('Problem 9.43') xlabel('Time, t (s)'); ylabel('Speed, v (ft/s)') axis([0 inf -inf infj) %inf self scales plot end % end of main function % THE DIFFERENTIAL EQUATION The Right Hand Side' function vdot = rhs(t,v,P,m,g) vdot = P/(m*v)-0.006*vA2/m; end % end of rhs ~ Page 6 / 6 Results from Matlab Code Problem 9.43 0 500 Time.t (s) 600 700 800 900 3) Acceleration is the slope of the velocity on the plot above. As time goes to infinity, the acceleration goes to zero. 4) As time goes to zero, the acceleration goes to infinity. This is why the initial velocity had to be inputted as a very small number (i.e. 0.001 ft/s) instead of zero. ...
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## This note was uploaded on 05/08/2008 for the course T&AM 203 taught by Professor Ruina during the Spring '08 term at Cornell University (Engineering School).

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Hw3 203 - 00 JD C 8 a ^ o o Q X ^ k) i 0 U 6 <T '? j* o...

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