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

Respuesta :

In the first second of movement, the body's position [tex]x[/tex] at time [tex]t[/tex] relative to the origin is

[tex]x=\left(27\dfrac{\rm m}{\rm s}\right)t[/tex]

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

[tex]x=\left(27\dfrac{\rm m}{\rm s}\right)(1\,\mathrm s)=27\,\mathrm m[/tex]

For every time [tex]t>1[/tex], its position is then given by

[tex]x=27\,\mathrm m+\left(27\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(-6\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]

so that after [tex]t=11[/tex] seconds, it will have undergone a displacement of

[tex]x=27\,\mathrm m+\left(27\dfrac{\rm m}{\rm s}\right)(11\,\mathrm s)+\dfrac12\left(-6\dfrac{\rm m}{\mathrm s^2}\right)(11\,\mathrm s)^2=\boxed{-39\,\mathrm m}[/tex]

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