Question

A particle's position is defined by: x = 4t^3-27t^2+17t+8, where x is expressed in meters and t in seconds. Determine (a) the total distance traveled the 2nd time the velocity is zero and (b) the velocity when the acceleration is zero.

Answer 1

Answer:

a)x = 100.56

b) t = 2.25 s

Explanation:

given equation

x = 4 t³ - 27 t² + 17 t + 8

to get velocity we need to differentiate the expression w.r.t time

$\dfrac{dx}{dt}=\dfrac{d}{dt}(4 t^3- 27 t^2+ 17 t + 8)$

on solving the differential equation

v = 12 t² -54 t + 17

at v = 0

using quadratic equation to solve

we get

$t = \dfrac{-(-54)\pm \sqrtr{-54^2-4\times 12\times 17}}{2\times 12}$

t = 4.16 s and 0.34 s

x = 4 × 4.16³ - 27 ×  4.16² + 17 × 4.16 + 8

x = -100.56

x = 100.56

b) v = 12 t² -54 t + 17

to get acceleration we need to differentiate the expression w.r.t time

$\dfrac{dv}{dt}=\dfrac{d}{dt}(12 t^2- 54 t+ 17)$

on solving the differential equation

a = 24 t -54

time at which acceleration will be zero

24 t  = 54

t = 2.25 s