Problem 19.11
The following data was collected for the distance travelled versus time for a rocket:
t, sec.
0
25
50
75
100
125
y, km
0
32
58
78
92
100
Use numerical differentiation to calculate velocity and acceleration at each time.
At time t=0, we can calculate the velocity by forward differences v=(32
‐
0)/(25
‐
0)=1.28 km/s. We can also
use a three point forward difference equation
(
2 )
4
(
)
3
( )
58
4
32
3
0
'( )
1.4
2
50
f
x
h
f
x
h
f
x
f
x
h
We can also fit a quintic polynomial to the data and take its derivative at zero
>> t=[0:25:125]
t =
0
25
50
75
100
125
>> y=[0 32 58 78 92 100]
y =
0
32
58
78
92
100
>> p=polyfit(t,y,5)
Warning: Polynomial is badly conditioned. Add points with distinct X
values, reduce the degree of the polynomial, or try centering
and scaling as described in HELP POLYFIT.
p =
‐
0.0000
0.0000
‐
0.0000
‐
0.0048
1.4000
0.0000
>> format short e
>> p
p =
‐
3.0478e
‐
022 1.0186e
‐
019
‐
1.2065e
‐
017
‐
4.8000e
‐
003 1.4000e+000 7.2060e
‐
015
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Which means that the data fits very well a quadratic polynomial
>> v0=p(5)
v0 =
1.4000e+000
We verify that a quadratic polynomial will do
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 Spring '09
 RAPHAELHAFTKA
 Numerical Analysis, Velocity, finite difference, forward differences

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