Acceleration to displacement

[mslena79]

1. two knights on horseback start from rest 88.0 m apart and ride directly toward each other to do battle. Sir George's acceleation has a magnitude of 0.300 m/s^2, Sir Alfred's has a magnitude of 0.200 m/s^2. Relative to Sir George's starting point, where do the knights collide?



2. a= (v-v0)/(t-t0), v=(x-x0)/(t-t0)



3. 0.300m/s^2=(v-0)/t, v=0.300m/s^2*t, 0.300m/s^2*t=(x-x0)/t, I don't know where to go from here!

 

[learningphysics]

You have a formula for displacement in terms of time and acceleration... try to use that to get the time when they collide...

 

[mslena79]

x=(0.300m/s^2 *t^2)/2 ?

 

[mslena79]

nevermind, I got it now!