course Phy231
3/3/10 11am
Determine the acceleration of an object whose velocity is initially 19 cm/s and which accelerates uniformly through a distance of 67 cm in 5.4 seconds.v0=19cm/s 'dt=67cm 'ds=5.4s vAve=67cm/5.4s=12.41cm/s 12.41cm/s=(19cm/s+vf) vf+19cm/s=24.8 vf=5.8cm/s
a=(5.8cm/s -19cm/s)/5.4s= -2.4cm/S^2
I tried using the formula 'ds=v0'dt+.5adt^2, but I didn't get the same answer after several tries. It does seem like this formula would be the easier way to find the acceleration.
I think I'm messing up when solving for a before I plug in the numbers. This is what I got... a=2('ds-v0'dt)/dt^2
You solved the equation correctly, and the result would agree with your previous result:
2 ( `ds - v0 `dt) / (`dt)^2 = 2 * (67 cm - 19 cm/s * 5.4 sec) / (5.4 sec)^2 = 2 * (-33 cm) / (30 sec^2) = -2.44 cm/s^2.