Question

A particle from rest accelerates the magnitude of 3 m/s^2. What is its distance after 8 seconds of travel?

Answer 1

$$d=v_0t+\dfrac{1}{2}at^2,$$ $$d=0+\dfrac{1}{2}\cdot3\ \dfrac{m}{s^2}\cdot(8\ s)^2=96\ m.$$