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 #51401 - 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

The interpolation polynomial in the Lagrange form is

$$V (t) = \sum_ {i = 0} ^ {n} V _ {i} * l _ {i} (t),$$

where $t_0 = 0, t_1 = 1, t_2 = 2, t_3 = 5, v_0 = 2, v_1 = 3, v_2 = 12, v_3 = 147$ .

Calculate

$$l _ {0} (t) = \prod_ {m = 1} ^ {3} \frac {t - t _ {m}}{t _ {0} - t _ {m}} = \frac {t - 1}{0 - 1} * \frac {t - 2}{0 - 2} * \frac {t - 5}{0 - 5} = - \frac {1}{1 0} (t ^ {3} - 8 t ^ {2} + 1 7 t - 1 0);$$ $$l _ {1} (t) = \prod_ {m = 0, m \neq 1} ^ {3} \frac {t - t _ {m}}{t _ {1} - t _ {m}} = \frac {t - 0}{1 - 0} * \frac {t - 2}{1 - 2} * \frac {t - 5}{1 - 5} = \frac {1}{4} (t ^ {3} - 7 t ^ {2} + 1 0 t);$$ $$l _ {2} (t) = \prod_ {m = 0, m \neq 2} ^ {3} \frac {t - t _ {m}}{t _ {2} - t _ {m}} = \frac {t - 0}{2 - 0} * \frac {t - 1}{2 - 1} * \frac {t - 5}{2 - 5} = - \frac {1}{6} (t ^ {3} - 6 t ^ {2} + 5 t);$$ $$l _ {3} (t) = \prod_ {m = 0,} ^ {2} \frac {t - t _ {m}}{t _ {3} - t _ {m}} = \frac {t - 0}{5 - 0} * \frac {t - 1}{5 - 1} * \frac {t - 2}{5 - 2} = \frac {1}{6 0} (t ^ {3} - 3 t ^ {2} + 2 t);$$

Velocity

$$\begin{array}{l} V (t) = - \frac {1}{1 0} * 2 * (t ^ {3} - 8 t ^ {2} + 1 7 t - 1 0) + \frac {1}{4} * 3 * (t ^ {3} - 7 t ^ {2} + 1 0 t) - \frac {1}{6} * 1 2 * \\ * (t ^ {3} - 6 t ^ {2} + 5 t) + \frac {1}{6 0} * 1 4 7 * (t ^ {3} - 3 t ^ {2} + 2 t) = t ^ {3} + t ^ {2} - t + 2; \\ \end{array}$$

Acceleration

$$a (t) = \frac {d V}{d t} = 3 t ^ {2} + 2 t - 1;$$

Acceleration at the end of 3 seconds

$$a(3) = 3 * 9 + 2 * 3 - 1 = 32;$$

The distance moved by a particle

$$S(t) = \int_{0}^{t} V(u) \, du;$$

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

$$S(3) = \int_{0}^{3} (t^3 + t^2 - t + 2) \, dt = \left(\frac{t^4}{4} + \frac{t^3}{3} - \frac{t^2}{2} + 2t\right) \Bigg|_{0}^{3} = 30.75;$$

(integration rule for powers and Newton-Leibnitz formula were applied here).

**Answer:** Distance = 30.75 and acceleration = 32.

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