Posted by **Susan** on Wednesday, October 5, 2011 at 9:08pm.

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?

- physics -
**drwls**, Thursday, October 6, 2011 at 12:17am
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 (t-0.21)^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

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