Question

Which of the following X-Y tables agrees withthe information in this problem?A runner moves 2.88 m/s north. She accelerates at0.350 m/s’ at a -52.0° angle. At the point in the motionwhere she is running directly east, how far is she fromher starting point?ХYB)XYC)хYA)ViVA2.88OV:O2.88V:oo2.88?V+0Via0.276 0.215a-0.276a0.350-0.3500.215PAx?- ?AX?Axt0ott

Answer 1

Given, initial velocity in the y-direction, vi=2.88 m/s

The acceleration of the runner, a=0.350 m/s²

The angle at which she is running, θ=-52.0°

The x-component of the acceleration is

`a_x=a*\cos \theta`

On substituting the known values,

`a_x=0.350*\cos -52.0^0=0.215m/s^2`

And the y-component of the acceleration is,

`a_y=a*\sin \theta`

On substituting the known values,

`a_y=0.350*\sin (-52.0)=-0.276m/s^2`

We can see that only table B has these values.

Therefore the correct answer is option B