1You are given data for the upward velocity of a rocket as a

(1)You are given data for the upward velocity of a rocket as a function of time in Table 1 below. Apply the direct interpolation method, use quadratic interpolant approximation of the velocity to find the acceleration at t= 16s.

(2)You are given data for the upward velocity of a rocket as a function of time in Table 1 below. Apply Netwon’s Divided Difference interpolation of the velocity, find the distance covered by the rocket from t=5s to t= 16s.

t(s)	0	10	15	20	22.5	30
v(t)(m/s)	0	227.04	362.75	517.35	602.97	901.67

Solution

(a) (1) Direct Interpolation Method:

v(t) = a₀ + a₁ t                                            (1)

To find the velocity at t = 16.

Choose t₀ = 15

and t₁ =20.

v(t₀) = 362.75

v(t₁) = 517.35

Substituting the values, we get:

v(15) = a₀ + a₁ (15) = 362.75             (2)

v(20) = a₀ + a₁ (20) = 517.35              (3)

(3) - (2) gives:

5 a₁ = 154.6

So, a₁ = 154.6/5

          = 30.92

Substituting in (1), we get

30.92 + 15 a₁ = 362.75

So, 15 a₁ = 362.75 - 30.92

               = 331.83

So, a₁ = 22.12.

Hence,

v(t) = a₀ + a₁ t

      = 30.92 + 331.83 t.

At t = 16,

v(16) = 30.92 + 331.83 x 16

        = 30.92 + 4992.45

         = 5023.37.

To find the acceleration, differentiate v(t) with respect to t:

Acceleration at time t is given by

a(t) = a₁ = 331.83.

(2) Use quadratic interpolant approximation:

v(t) = a₀ + a₁ t + a₂ t².

Take

t₀ = 10, v(t₀) = 227.04

t₁ = 15, v(t₁) = 362.75

t₂ = 20, v(t₂) = 517.75.

So, we get:

v(10) = a₀ + a₁ (10) + a₂ (10)²

v(15) = a₀ + a₁ (15) + a₂ (15)²

v(20) = a₀ + a₁ (20) + a₂ (20)²

Solving these 3 equations, we get,

a₀ = 12.05

a₁ = 17.733

a₂ = 0.3766.

Hence

v(t) = 12.05 + 17.733 t + 0.3766 t ².

Differentiating, we get acceleration at t as:

a(t) = 17.733 + 0.7532 t.

At t=16, substitute for t = 16. you get the acceleration.

As per Directions for Answering, only (a) is answered, since it is a long question.