Question

Given the following table of velocity data 1, 0 05 10 15 20 25 3.0 3.5 4.0 V m/s 0 12 16 1.4 2.0 2.0 1.8 1.6 1.3 a) Estimate
0 0
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Answer #1
x f(x)
0 0
0.5 1.2
1.0 1.6
1.5 1.4
2.0 2.0
2.5 2.0
3.0 1.8
3.5 1.6
4.0 1.3

(1) Using Newton's forward/backward differentiation method to find solution

[dV/dt] at (t = t0 )=1/h⋅(ΔY0-1/2⋅Δ2Y0+1/3⋅Δ3Y0-1/4⋅Δ4Y0)

put h = 0.5

and value of  ΔY0, Δ2Y0 , Δ3Y0 , Δ4Y0  accordingly.

We will get

dV/dt = 1.333333 at t=1.5

{\color{Red} }(4)

for acceleration at t = 4

using Newton's backward differentiation table is as follows.

[dVd/t] (t=tn)=1/h⋅(∇Yn+ 1/2⋅∇2Yn+1/3⋅∇3Yn+1/4⋅∇4Yn)

pt h = 0.5 and value of

∇Yn,2Yn3Yn,4Yn accordingly

we will get

dVd/t = -0.91667 at t = 4

(2) using newtons simpson 1/3 rule rule

location at t = 4 is X = 6.15

(3)  using newtons simpson 1/3 rule rule

  location at t = 1.5 is X = 1.56667

You may ask further if you have more doubts

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Thank you


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