The position as a function of time xt yt of a squirrel runni

The position as a function of time [x(t), y(t)] of a squirrel running on a grass field is given by (in meters), x(t) = -0.28t^2: + 6.5t + 61 and y(t) = 0.18t^2 - 8,5t +65. The length of the position vector is: r(t) = and the angle of the position vector is: For 0 lessthanorequalto t lessthanorequalto 30 s, plot the trajectory (position) of the squirrel as y versus x, the length of the position vector r(t) versus t, and the angle (in degrees) versus t in three different plots in one column. Include labels, title, and grid lines for each plot.

Solution

I don\'t have MATLAB installed on my computer right now, so this is untested. It should be pretty close though.

clear all;

t=0:0.1:30;

for i=1:length(t)
%x position
x(i)=-.28t(i)^2+6.5t(i)+61;
%y position
y(i)=.18t(i)^2-8.5t(i)+65;
%length of the position vector
length(i)=(x(i)^2+y(i)^2)^(0.5);
%angle of the vector
angle(i)=(180/pi)*atan(y(i)/x(i));
end

figure(1)
plot(x,y)
xlabel(\'x\')
ylabel(\'y\')
title(\'part a, position\')

figure(2)
plot(t,length)
xlabel(\'t\')
ylabel(\'vector length\')
title(\'part b, length of the position vector\')

figure(3)
plot(t,angle)
xlabel(\'t\')
ylabel(\'angle [deg]\')
title(\'part c, angle of the position vector\')