Question

A boat takes off from a dock at 2.5 m/s and speeds up at 4.2 m/s squared for six seconds how far has the most traveled

Answer 1

The boat's position
`x` relative to its starting point
`x_0=0` is determined by

`x=x_0+v_0t+\frac12at^2`

where
`v_0` is its initial velocity,
`a` is its acceleration, and
`t` is time. After
`t=6\,\mathrm s`, the boat has traveled

`x=\left(2.5\,(\mathrm m)/(\mathrm s)\right)(6\,\mathrm s)+\frac12\left(4.2\,(\mathrm m)/(\mathrm s^2)\right)(6\,\mathrm s)^2`

`\implies x=91\,\mathrm m`