The position of a particle is given by the function x = \left(2t^3 - 6t^2 + 12\right) m, where t is in s. Question:At what time is the acceleration zero? (with working out)

Explanation:

Given the position of a particle  expressed by the equation x = (2t^3 - 6t^2 + 12)m, where t is in seconds, the acceleration function can be gotten by taking the second derivative of the function with respect to t as shown;

a = d/dt(dx/dt)

First let us get dx/dt

dx/dt = 3(2)t³⁻¹-2(6)t²⁻¹+0

dx/dt = 6t²-12t

a = d/dt(dx/dt)

a = d/dx(6t²-12t)

a = 2(6)t²⁻¹-12t¹⁻¹

a = 12t - 12t⁰

a = 12t-12

If the acceleration is zero, then;

12t-12 = 0

add 12 to both sides

12t-12+12 = 0+12

12t = 12

t = 12/12

t = 1sec

Hence the time when acceleration is zero is 1sec

in order to follow the instructions and complete the statements, i need to use data from table c of your student guide. can you think of a way to let me see your student guide for a while ? i can't answer the question without the data in table c.