A particle moves along the x-axis according to x = t⁴ - 8t³ + 18t². The velocity at the first positive instant when its acceleration becomes zero is:
Options

Question

A particle moves along the x-axis according to x = t⁴ − 8t³ + 18t². The velocity at the first positive instant when its acceleration becomes zero is:

G1 Physics · Accepted Answer

v = dx/dt = 4t³ − 24t² + 36t = 4t(t−3)². a = dv/dt = 12t² − 48t + 36 = 12(t−1)(t−3). The first positive instant for a=0 is t=1 s. Therefore v(1)=4(1)(1−3)²=16 m s⁻¹.