Question

An object moves in along the x-axis with an acceleration given by: a = 5t (m/s2). The position of the object at t=0.0 is 6 m, and its velocity at t=0.0 s is 4 m/s. Calculate the position at t=5 s.​

Answer 1

The position of the object at
`t = 5\,s` is 130.167 meters.

Step-by-step explanation:

Let
`a(t) = 5\cdot t\,\left[(m)/(s^(2)) \right]` 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

`v(t) = \int {a(t)} \, dt` (1)

`v(t) = 5\int {t} \, dt`

`v(t) = (5)/(2)\cdot t^(2)+v_(o)` (2)

Where
`v_(o)` is the initial velocity of the object, measured in meters per second.

Position

`s(t) = \int {v(t)} \, dt` (3)

`s(t) = (5)/(2)\int {t^(2)} \, dt+v_(o)\int \, dt`

`s(t) = (5)/(6)\cdot t^(3)+v_(o)\cdot t + s_(o)` (4)

Where
`s_(o)` is the initial position of the object, measured in meters per second.

If we know that
`s_(o) = 6\,m`,
`v_(o) = 4\,(m)/(s)` and
`t = 5\,s`, then the position of the object is:

`s(5) = (5)/(6)\cdot (5)^(3)+\left(4\right)\cdot (5)+6`

`s(5) = 130.167\,m`

The position of the object at
`t = 5\,s` is 130.167 meters.