Answer:
The box displacement after 6 seconds is 66 meters.
Explanation:
Let suppose that velocity given in statement represents the initial velocity of the box and, likewise, the box accelerates at constant rate. Then, the displacement of the object ([tex]\Delta s[/tex]), in meters, can be determined by the following expression:
[tex]\Delta s = v_{o}\cdot t+\frac{1}{2}\cdot a\cdot t^{2}[/tex] (1)
Where:
[tex]v_{o}[/tex] - Initial velocity, in meters per second.
[tex]t[/tex] - Time, in seconds.
[tex]a[/tex] - Acceleration, in meters per square second.
If we know that [tex]v_{o} = 5\,\frac{m}{s}[/tex], [tex]t = 6\,s[/tex] and [tex]a = 2\,\frac{m}{s^{2}}[/tex], then the box displacement after 6 seconds is:
[tex]\Delta s = 66\,m[/tex]
The box displacement after 6 seconds is 66 meters.