Question

The time versus velocity data of a particle is given in the table below. Use Lagrange’s
interpolation formula to find the distance moved by a particle and its acceleration at
the end of 3 seconds.
t: 0, 1, 2, 5
v: 2, 3, 12, 147

Answer 1

Answer on Question #51465 – Math – Algorithms | Quantitative methods

The time versus velocity data of a particle is given in the table below. Use Lagrange's interpolation formula to find the distance moved by a particle and its acceleration at the end of 3 seconds.

t: 0, 1, 2, 5

v: 2, 3, 12, 147

Solution

Lagrange's interpolation formula:

$$L(x) = \sum_{i=0}^{n} f_i \prod_{\substack{0 \leq k \leq n \\ k \neq i}}^{n} \frac{x - x_k}{x_i - x_k}$$

In this problem:

Velocity

$$L(t) = \sum_{i=0}^{3} v_i \prod_{\substack{0 \leq k \leq n \\ k \neq i}}^{3} \frac{t - t_k}{t_i - t_k} = 2 \frac{(t-1)(t-2)(t-5)}{(-1)*(-2)*(-5)} + 3 \frac{t(t-2)(t-5)}{1*(-1)*(-4)} + 12 \frac{t(t-1)(t-5)}{2*1*(-3)} + 147 \frac{t(t-1)(t-2)}{5*4*3} = t^3 + t^2 - t + 2.$$

The distance moved by a particle at the end of 3 seconds

$$\begin{array}{l} \text{So, } S(3) = \int_{0}^{3} v(t) dt = \left(\frac{1}{4} t^4 + \frac{1}{3} t^3 - \frac{1}{2} t^2 + 2t\right) \big|_{t=0}^{t=3} = \left(\frac{1}{4} 3^4 + \frac{1}{3} 3^3 - \frac{1}{2} 3^2 + 2 \cdot 3\right) - \\ - \left(\frac{1}{4} 0^4 + \frac{1}{3} 0^3 - \frac{1}{2} 0^2 + 2 \cdot 0\right) = \frac{123}{4} = 30.75. \end{array}$$

Acceleration

$$a = \frac{dv}{dt} = 3t^2 + 2t - 1;$$

Acceleration at the end of 3 seconds

$$a(3) = 32.$$

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