Question

A body moving in the positive x direction passes the origin at time t=0.Between t=0 and t=1second,the body has a constant speed of 27 meters per second. At t = 1 second, the body is given a constant acceleration of 6 meters per second squared in the negative x direction. The position x of the body at t = 11 seconds is

Answer 1

In the first second of movement, the body's position
`x` at time
`t` relative to the origin is

`x=\left(27(\rm m)/(\rm s)\right)t`

so that after the first second, it will have undergone a displacement of

`x=\left(27(\rm m)/(\rm s)\right)(1\,\mathrm s)=27\,\mathrm m`

For every time
`t>1`, its position is then given by

`x=27\,\mathrm m+\left(27(\rm m)/(\rm s)\right)t+\frac12\left(-6(\rm m)/(\mathrm s^2)\right)t^2`

so that after
`t=11` seconds, it will have undergone a displacement of

`x=27\,\mathrm m+\left(27(\rm m)/(\rm s)\right)(11\,\mathrm s)+\frac12\left(-6(\rm m)/(\mathrm s^2)\right)(11\,\mathrm s)^2=\boxed{-39\,\mathrm m}`

so it ends up 39 m to the left of where it started (taking the right of the origin to be the positive
`x` direction).