Respuesta :

The runner has initial velocity

[tex]\vec v_0=\left(2.38\,\dfrac{\rm m}{\rm s}\right)\,\vec\jmath[/tex]

and acceleration

[tex]\vec a=\left(0.310\,\dfrac{\rm m}{\mathrm s^2}\right)\left(\cos(-62.0^\circ)\,\vec\imath+\sin(-62.0^\circ)\,\vec\jmath\right)[/tex]

Her velocity [tex]\vec v[/tex] at time [tex]t[/tex] is

[tex]\vec v=\vec v_0+\vec at[/tex]

[tex]\vec v=\left(0.310\,\dfrac{\rm m}{\mathrm s^2}\right)\cos(-62.0^\circ)\,t\,\vec\imath+\left(2.38\,\dfrac{\rm m}{\rm s}+\left(0.310\,\dfrac{\rm m}{\mathrm s^2}\right)\sin(-62.0^\circ)\,t\right)\,\vec\jmath[/tex]

The velocity vector will be pointing directly east when when the vertical component of this vector is 0 and the horizontal component has a positive sign. The second condition is always true, since its value is a constant [tex]0.146\,\dfrac{\rm m}{\rm s}[/tex]. The first condition is met at

[tex]2.38\,\dfrac{\rm m}{\rm s}+\left(0.310\,\dfrac{\rm m}{\mathrm s^2}\right)\sin(-62.0^\circ)\,t=0\implies t=8.70\,\rm s[/tex]