Question

A particle starts at time at the position The velocity of the particle is written in the polar basis associated with its current position, and is: Matlab/Mathematica input: x0 = 13 y0 = -12 What is the position of at ?

A particle P starts at time t=0 s at the position x = 13 m y = –12 m. The velocity ✓ of the particle is written in the polar basis associated with its current position, and is: ū = [-3, – 3] m/s. Matlab/Mathematica input: x0 = 13 yo = -12 What is the position of P at t = 3 s? X = m, y= m

Answer 1

Analytical solution V = dà Sdxaluat where, x is position OC= vtt to at t V is a velocity t is time to position at t Vo vy o V = 633, 34 ms. to = 15m Yo = y2m. . 0 X = Vx. t t 40 a = -3+13. t=35 2=-9+13 = @=4 m) when to 35 y= Vyt+ % y=-3t-12 = -4-2 (t - 3) [Y=~21 m) when t=35 For matlab J-U[+]*[:] -TX)=[V][+]+[Xo]

x Chegg1.m Bisection.m x VelocityChegg. m x + Name Value [4;-21) (13;-12] DO 3- Differentiation.m x RootlocusChegg1.m x Untitled2.m clear all x0 = input('Enter x0 = \n'); y0 = input('Enter yo = \n'); t = input('time = \n'); DO = (x0;y0]; V = (-3;-3]; 7 T = [t]; D = V*T + DO; x = D(1) 10 - y = D(2) (-3;-3] IIIIIIIIII Command Window >> VelocityChegg Enter x0 = 13 Enter yo = -12 time = X = -21 fx >> |

MatLab Code:

clear all
x0 = input('Enter x0 = \n') ;
y0 = input('Enter y0 = \n');
t = input('time = \n');
D0 = [x0;y0];
V = [-3;-3];
T = [t];
D = V*T + D0;

x = D(1)
y = D(2)

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Answer:

X = 4m

Y = -12m

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Thank you..