Answer:
The position of the object at [tex]t = 5\,s[/tex] is 130.167 meters.
Explanation:
Let [tex]a(t) = 5\cdot t\,\left[\frac{m}{s^{2}} \right][/tex] the acceleration experimented by the object along the x-axis. We obtain the equation for the position of the object by integrating in acceleration formula twice:
Velocity
[tex]v(t) = \int {a(t)} \, dt[/tex] (1)
[tex]v(t) = 5\int {t} \, dt[/tex]
[tex]v(t) = \frac{5}{2}\cdot t^{2}+v_{o}[/tex] (2)
Where [tex]v_{o}[/tex] is the initial velocity of the object, measured in meters per second.
Position
[tex]s(t) = \int {v(t)} \, dt[/tex] (3)
[tex]s(t) = \frac{5}{2}\int {t^{2}} \, dt+v_{o}\int \, dt[/tex]
[tex]s(t) = \frac{5}{6}\cdot t^{3}+v_{o}\cdot t + s_{o}[/tex] (4)
Where [tex]s_{o}[/tex] is the initial position of the object, measured in meters per second.
If we know that [tex]s_{o} = 6\,m[/tex], [tex]v_{o} = 4\,\frac{m}{s}[/tex] and [tex]t = 5\,s[/tex], then the position of the object is:
[tex]s(5) = \frac{5}{6}\cdot (5)^{3}+\left(4\right)\cdot (5)+6[/tex]
[tex]s(5) = 130.167\,m[/tex]
The position of the object at [tex]t = 5\,s[/tex] is 130.167 meters.
