physics
posted by Susan on .
Tom & Dick are 25 m apart at rest. Tom accelerates at 5 m/s/s towards Dick. 0.21 sec later Dick accelerates 6 m/s/s towards Tom. Where do they meet, how fast is each traveling when they meet, and when do they meet?

Write equations for Tom and Dick's locations and solve for the time they are equal. After you have the time, the other two unknowns are easily computed.
X1 = (5/2) t^2 is Tom's location in feet, if t is in seconds. Assume he starts at the origin, X = 0
X2 = 25  3 (t0.21)^2 is Dick's location for t>0.21 s
Solve for t when X1 = X2.
Tom's velocity at that time is 5 t ft/s