Problem 1. You recorded the following measurements for a moving object as follow

Question

Problem 1. You recorded the following measurements for a moving object as following. Estimate the velocity of the object at t -1.0 sec,t-2.4 sec and t-3.5 sec. 0.0 0.6 7 2.0 2.5 Time (sec) 3.2 121 132140 105 157 Velocity (m/s) 90 a) Use linear interpolation to estimate the velocity of the object "by hand". If you are unable to interpolate the velocity for any of the t values, explain why. b) Use MATLAB and linear interpolation to estimate the velocities c) Use MATLAB and spline interpolation to estimate the velocities d) Use MATLAB "polyfit" command to find the polynomial function that passes all points. Clearly state the function! e) Use polynomial interpolation function from d) to estimate the velocities f) Plot these three interpolation curves for the domain0.0 3.2] Mark' on these six data points. Label all axes and show legends

Explanation / Answer

a)

By linear interpolation velocity is given by formula

v(t)=a0+a1t

We need velocity at t=1s, t= 2.4s t= 3.5s using above table

For t=1

let us choose closest point of this time in between which t=1 falls so we get

t0=0.6 and t1= 1.7

Therfore, t0=0.6 v( t0) =132

t1= 1.7 v(t1)=140

which gives us equations

v(0.6)=a0 +a1(0.6) = 132

v(1.7)=a0 +a1(1.7) = 140

Let us write equation in matrix

1 0.6 a0 132

1 1.7 a1 = 140

By solving equations we get

a0=-127.63 a1=7.28

hence

v(t) = a0+a1 t

= -127.63 +7.28 x t

for t=1

= -127.63 +7.28 x 1

= -120.35

Similarly we can calculate for t=2.4

let us choose closest point of this time in between which t=1 falls so we get

t0=2.0 and t1= 2.5

Therefore, t0=2.0 v( t0) =105

t1= 2.5 v(t1)=157

which gives us equations

v(2.0)=a0 +a1(2.0) = 105

v(2.5)=a0 +a1(2.5) = 157

Let us write equation in matrix

1 2.0 a0 105

1 2.5 a1 = 157

By solving equations we get

a0=-103 a1=104

hence

v(t) = a0+a1 t

= -103 +(104 x t)

for t=2.4

= -103 + 104x 2.4

= -105

For t=3.5 we cannot interpolate as upper value is not available as last being t=3.2 which is less than 3.5 and any value bigger than 3.5 is not avalaible so interpolation is not possible

b) For t=1 and t=2.4

x=[0.0 0.6 1.7 2.0 2.5 3.2]
y=[121 132 140 105 157 90]
x1=0:1:3.2
y1=interp1(x,y,x1,'linear');
y1
x2=0:2.4:3.2
y2=interp1(x,y,x2,'linear');
y2

c) x3=0:1:3.2
y3=interp1(x,y,x3,'spline');
y3
x4=0:2.4:3.2
y4=interp1(x,y,x4,'spline');
y4

d). a=polyfit(x,y,1) ro get simple 1 degree linear regression

e)

x5=0:1:3.2
y5 = polyval(polyfit(x,y,1),x5);
y5
x6=0:2.4:3.2
y6 = polyval(polyfit(x,y,1),x6);
y6

f)

plot(x1,y1)
hold
plot(x3,y3)
plot(x5,y5)
title('Velocity ','fontsize',20)
ylabel('Meters per sec','fontsize',15)
xlabel('Time(seconds)','fontsize',15)
legend('Linear Interpolation','Spline Interpolation', 'Polynomial Interpolation')

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