Question

A runner moves 2.88 m/s north. She accelerates at 0.350 m/s^2 at a -52.0 angle. At the point in the motion where she is running directly east, what is Δx?

Answer 1

The runner has initial velocity vector

$\vec v_0=\left(2.88\dfrac{\rm m}{\rm s}\right)\,\vec\jmath$

and acceleration vector

$\vec a=\left(0.350\dfrac{\rm m}{\mathrm s^2}\right)(\cos(-52.0^\circ)\,\vec\imath+\sin(-52.0^\circ)\,\vec\jmath)$

so that her velocity at time $t$ is

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

She runs directly east when the vertical component of $\vec v$ is 0:

$2.88\dfrac{\rm m}{\rm s}+\left(0.350\,\dfrac{\rm m}{\mathrm s^2}\right)\sin(-52.0^\circ)\,t=0\implies t=10.4\,\rm s$

It's not clear what you're supposed to find at this particular time... possibly her position vector? In that case, assuming she starts at the origin, her position at time $t$ would be

$\vec x=\vec v_0t+\dfrac12\vec at^2$

so that after 10.4 s, her position would be

$\vec x=(10.1\,\mathrm m)\,\vec\imath+(17.2\,\mathrm m)\,\vec\jmath$

which is 19.9 m away from her starting position.

Answer 2

Answer:

Δx = 11.7 and Δy = 15