Question

Please please please i really need to know

Answer 1

The runner has initial velocity

$\vec v_0=\left(2.38\,\dfrac{\rm m}{\rm s}\right)\,\vec\jmath$

and acceleration

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

Her velocity $\vec v$ at time $t$ is

$\vec v=\vec v_0+\vec at$

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

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 $0.146\,\dfrac{\rm m}{\rm s}$. The first condition is met at

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