EBK NUMERICAL METHODS FOR ENGINEERS
EBK NUMERICAL METHODS FOR ENGINEERS
7th Edition
ISBN: 9780100254145
Author: Chapra
Publisher: YUZU
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Differential Equations
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Chapter 24, Problem 44P

A jet fighter's position on an aircraft carrier's runway was timed during landing:

t, s 0 0.52 1.04 1.75 2.37 3.25 3.83
x, m 153 185 208 249 261 271 273

where x is the distance from the end of the carrier. Estimate (a) velocity ( d x / d t ) and (b) acceleration ( d v / d t ) using numerical differentiation.

(a)

Expert Solution
Check Mark
To determine

To calculate: The velocity of the jet fighter landing on an aircraft carrier’s runaway using Numerical Integration.

Answer to Problem 44P

Solution:

The velocities at different points of time of a Jet fighter are given below,

t,s 0 0.52 1.04 1.75 2.37 3.25 3.83
x,m 153 185 210 249 261 271 273
v,m/s 68.26923 54.80769 50.97398 35.93855 16.05180 6.59273 0.303817

Explanation of Solution

Given Information:

The position of a Jet fighter on an aircraft carrier’s runway is,

t,s 0 0.52 1.04 1.75 2.37 3.25 3.83
x,m 153 185 210 249 261 271 273

Formula Used:

Second-order Lagrange interpolating polynomial.

f′(t)=f(ti−1)2t−ti−ti+1(ti−1−ti)(ti−1−ti+1)+f(ti)2t−ti−1−ti+1(ti−ti−1)(ti−ti+1)+f(ti+1)2t−ti−1−ti(ti+1−ti−1)(ti+1−ti)…… (1)

Here, t is the value at which the derivative is to be estimated.

Calculation:

Consider the function f(x) to be position (x) as shown in table above. So, f′(x) will represent the acceleration (v).

Apply second-order Lagrange interpolating polynomial to each set of three adjacent points for calculation of velocities at different points.

Calculate f′(0).

Substitute t=0 and i=1 in equation(1).

f′(0)=f(0)2(0)−t1−t2(t0−t1)(t0−t2)+f(t1)2(0)−t0−t2(t1−t0)(t1−t2)+f(t2)2(0)−t0−t1(t2−t0)(t2−t1)

Substitute the value of function from above table.

f′(0)=153(2(0)−0.52−1.04(0−0.52)(0−1.04))+185(2(0)−0−1.04(0.52−0)(0.52−1.04))+210(2(0)−0−0.52(1.04−0)(1.04−0.52))=68.26923

Calculate f′(0.52).

Substitute t=0.52 and i=2 in equation(1).

f′(0.52)=f(0.52)2(0.52)−t2−t3(t1−t2)(t1−t3)+f(t2)2(0.52)−t1−t3(t2−t1)(t2−t3)+f(t3)2(0.52)−t1−t2(t3−t1)(t3−t2)

Substitute the value of function from above table.

f′(0.52)=153(2(0.52)−0.52−1.04(0−0.52)(0−1.04))+185(2(0.52)−0−1.04(0.52−0)(0.52−1.04))+210(2(0.52)−0−0.52(1.04−0)(1.04−0.52))=54.80769

Calculate f′(1.04).

Substitute t=1.04 and i=3 in equation(1).

f′(1.04)=f(1.04)2(1.04)−t3−t4(t2−t3)(t2−t4)+f(t3)2(1.04)−t2−t4(t3−t2)(t3−t4)+f(t4)2(1.04)−t2−t3(t4−t2)(t4−t3)

Substitute the value of function from above table.

f′(1.04)=185(2(1.04)−1.04−1.75(0.52−1.04)(0.52−1.75))+210(2(1.04)−0.52−1.75(1.04−0.52)(1.04−1.75))+249(2(1.04)−0.52−1.04(1.75−0.52)(1.75−1.04))=50.97398

Calculate f′(1.75).

Substitute t=1.75 and i=1 in equation(1).

f′(1.75)=f(1.75)2(1.75)−t4−t5(t3−t4)(t3−t5)+f(t4)2(1.75)−t3−t5(t4−t3)(t4−t5)+f(t5)2(1.75)−t4−t3(t5−t3)(t5−t4)

Substitute the value of function from above table.

f′(1.75)=210(2(1.75)−1.75−2.37(1.04−1.75)(1.04−2.37))+249(2(1.75)−1.04−2.37(1.75−1.04)(1.75−2.37))+261(2(1.75)−1.04−1.75(2.37−1.04)(2.37−1.75))=35.93855

Calculate f′(2.37).

Substitute t=2.37 and i=5 in equation(1).

f′(2.37)=f(2.37)2(2.37)−t5−t6(t4−t5)(t4−t6)+f(t5)2(2.37)−t4−t6(t5−t6)(t5−t4)+f(t6)2(2.37)−t4−t5(t6−t4)(t6−t5)

Substitute the value of function from above table.

f′(2.37)=249(2(2.37)−2.37−3.25(1.75−2.37)(1.75−3.25))+261(2(2.37)−1.75−3.25(2.37−1.75)(2.37−3.25))+271(2(2.37)−1.75−2.37(3.25−1.75)(3.25−2.37))=16.05180

Calculate f′(3.25).

Substitute t=3.25 and i=6 in equation(1).

f′(3.25)=f(3.25)2(3.25)−t6−t7(t5−t6)(t5−t7)+f(t6)2(3.25)−t6−t7(t6−t5)(t6−t7)+f(t7)2(3.25)−t5−t6(t7−t6)(t7−t5)

Substitute the value of function from above table.

f′(3.25)=261(2(3.25)−3.25−3.83(2.37−3.25)(2.37−3.83))+271(2(3.25)−2.37−3.83(3.25−2.37)(3.25−3.83))+273(2(3.25)−2.37−3.25(3.83−2.37)(3.83−3.25))=6.59273

Calculate f′(3.83).

Substitute t=3.83 and i=7 in equation(1).

f′(3.83)=f(3.83)2(3.83)−t7−t8(t6−t7)(t6−t8)+f(t7)2(3.83)−t6−t8(t7−t6)(t7−t8)+f(t8)2(3.83)−t6−t8(t8−t6)(t8−t7)

Substitute the value of function from above table.

f′(3.83)=261(2(3.83)−3.25−3.83(2.37−3.25)(2.37−3.83))+271(2(3.83)−2.37−3.83(3.25−2.37)(3.25−3.83))+273(2(3.83)−2.37−3.25(3.83−2.37)(3.83−3.25))=0.303817

The velocities at different points of time of a Jet fighter are given below,

t,s 0 0.52 1.04 1.75 2.37 3.25 3.83
x,m 153 185 210 249 261 271 273
v,m/s 68.26923 54.80769 50.97398 35.93855 16.05180 6.59273 0.303817

(b)

Expert Solution
Check Mark
To determine

To calculate: The acceleration of the jet fighter landing on an aircraft carrier’s runaway using Numerical Integration.

Answer to Problem 44P

Solution:

The complete table of acceleration for different values of velocity is given below,

t,s x,m v=dxdt a=dvdt
0 153 68.26923 -35.14510
0.52 185 54.80769 -16.63004
1.04 210 50.97398 -13.20841
1.75 249 35.93855 -26.99478
2.37 261 16.05180 -23.26046
3.25 271 6.59273 -10.80560
3.83 273 0.303817 -10.88029

Explanation of Solution

Given Information:

The velocities at different points of time of a Jet fighter are given below as calculated in Part (a).

t,s 0 0.52 1.04 1.75 2.37 3.25 3.83
x,m 153 185 210 249 261 271 273
v,m/s 68.26923 54.80769 50.97398 35.93855 16.05180 6.59273 0.303817

Formula Used:

Second-order Lagrange interpolating polynomial.

f′(t)=f(ti−1)2t−ti−ti+1(ti−1−ti)(ti−1−ti+1)+f(ti)2t−ti−1−ti+1(ti−ti−1)(ti−ti+1)+f(ti+1)2t−ti−1−ti(ti+1−ti−1)(ti+1−ti)…… (1)

Here, t is the value at which the derivative is to be estimated.

Calculation:

Consider the function f(x) to be velocity (v) as shown in table above. So, f′(x) will represent the acceleration (a).

Calculate the acceleration (a) at different points with respect to the velocities (v) given in the above table with the help of Equation (1).

Calculate f′(0).

Substitute t=0 and i=1 in equation(1).

f′(0)=f(0)2(0)−t1−t2(t0−t1)(t0−t2)+f(t1)2(0)−t0−t2(t1−t0)(t1−t2)+f(t2)2(0)−t0−t1(t2−t0)(t2−t1)

Substitute the value of function from above table.

f′(0)=68.26923(2(0)−0.52−1.04(0−0.52)(0−1.04))+54.80769(2(0)−0−1.04(0.52−0)(0.52−1.04))+50.97398(2(0)−0−0.52(1.04−0)(1.04−0.52))=−35.14510

Calculate f′(0.52).

Substitute t=0.52 and i=2 in equation(1).

f′(0.52)=f(0.52)2(0.52)−t2−t3(t1−t2)(t1−t3)+f(t2)2(0.52)−t1−t3(t2−t1)(t2−t3)+f(t2)2(0.52)−t1−t2(t3−t1)(t3−t2)

Substitute the value of function from above table.

f′(0.52)=68.26923(2(0.52)−0.52−1.04(0−0.52)(0−1.04))+54.80769(2(0.52)−0−1.04(0.52−0)(0.52−1.04))+50.97398(2(0.52)−0−0.52(1.04−0)(1.04−0.52))=−16.63004

Calculate f′(1.04).

Substitute t=1.04 and i=3 in equation(1).

f′(1.04)=f(1.04)2(1.04)−t3−t4(t2−t3)(t2−t4)+f(t1)2(1.04)−t2−t4(t3−t2)(t3−t4)+f(t4)2(1.04)−t2−t3(t4−t2)(t4−t3)

Substitute the value of function from above table.

f′(1.04)=54.80769(2(1.04)−1.04−1.75(0.52−1.04)(0.52−1.75))+50.97398(2(1.04)−0.52−1.75(1.04−0.52)(1.04−1.75))+35.93855(2(1.04)−0.52−1.04(1.75−0.52)(1.75−1.04))=−13.20841

Calculate f′(1.75).

Substitute t=1.75 and i=4 in equation(1).

f′(1.75)=f(1.75)2(1.75)−t4−t5(t3−t4)(t3−t5)+f(t1)2(1.75)−t3−t5(t4−t3)(t4−t5)+f(t2)2(1.75)−t4−t3(t5−t3)(t5−t4)

Substitute the value of function from above table.

f′(1.75)=50.97398(2(1.75)−1.75−2.37(1.04−1.75)(1.04−2.37))+35.93855(2(1.75)−1.04−2.37(1.75−1.04)(1.75−2.37))+16.05180(2(1.75)−1.04−1.75(2.37−1.04)(2.37−1.75))=−26.99478

Calculate f′(2.37).

Substitute t=2.37 and i=1 in equation(1).

f′(2.37)=f(2.37)2(2.37)−t1−t2(t0−t1)(t0−t2)+f(t1)2(2.37)−t0−t2(t1−t0)(t1−t2)+f(t2)2(2.37)−t0−t1(t2−t0)(t2−t1)

Substitute the value of function from above table.

f′(2.37)=35.93855(2(2.37)−2.37−3.25(1.75−2.37)(1.75−3.25))+16.05180(2(2.37)−1.75−3.25(2.37−1.75)(2.37−3.25))+6.59273(2(2.37)−1.75−2.37(3.25−1.75)(3.25−2.37))=−23.26046

Calculate f′(3.25).

Substitute t=3.25 and i=1 in equation(1).

f′(3.25)=f(3.25)2(3.25)−t6−t7(t5−t6)(t5−t7)+f(t6)2(3.25)−t6−t7(t6−t5)(t6−t7)+f(t7)2(3.25)−t5−t6(t7−t6)(t7−t5)

Substitute the value of function from above table.

f′(3.25)=16.05180(2(3.25)−3.25−3.83(2.37−3.25)(2.37−3.83))+6.59273(2(3.25)−2.37−3.83(3.25−2.37)(3.25−3.83))+0.303817(2(3.25)−2.37−3.25(3.83−2.37)(3.83−3.25))=−10.80560

Calculate f′(3.83).

Substitute t=3.83 and i=7 in equation(1).

f′(3.83)=f(3.83)2(3.83)−t7−t8(t6−t7)(t6−t8)+f(t7)2(3.83)−t6−t8(t7−t6)(t7−t8)+f(t8)2(3.83)−t6−t8(t8−t6)(t8−t7)

Substitute the value of function from above table.

f′(3.83)=16.05180(2(3.83)−3.25−3.83(2.37−3.25)(2.37−3.83))+6.59273(2(3.83)−2.37−3.83(3.25−2.37)(3.25−3.83))+0.303817(2(3.83)−2.37−3.25(3.83−2.37)(3.83−3.25))=−10.88029

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Chapter 24 Solutions

EBK NUMERICAL METHODS FOR ENGINEERS

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