# The Distance the Particle Travels with Given Position Function x(t) = t^4 - 8t^2

**Problem**

Given the position function *x*(*t*) = *t*^{4} - 8*t*^{2}, find the distance that the particle travels at *t* = 0 to *t* = 4.

A. 160 | C. 140 |

B. 150 | D. 130 |

**Answer Key**

[ A ]

**Solution**

$\displaystyle s = \int_{t_1}^{t_2} \sqrt{1 + \left( \dfrac{dx}{dt} \right)^2} \, dt$

$\displaystyle s = \int_0^4 \sqrt{1 + (4t^3 - 16t)^2} \, dt$

$s = 160.9$ ← Answer: [ A ]

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