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October 24, 2014

October 24, 2014

Posted by **Anonymous** on Sunday, October 7, 2007 at 10:56pm.

I understnad that bike 1 passes at t=o and 10 sec later bike 2 passes. I just don't know what formula to use to help me figure out the problem

- physics -
**Quidditch**, Sunday, October 7, 2007 at 11:14pmFirst I would suggest converting mph to feet/second for both riders.

When the second rider passes you, the first rider has been riding for 10 seconds past you. Find that distance (10 seconds * first rider's speed in feet/second). The second rider is catching up to the first rider at a speed of 10 mph (convert to ft/sec). Find how many seconds it will take for the second rider to catch the first rider. Multiply this time by the ft/second of the second rider to find how far he (and the first rider) have gone past your position when the second rider cathes the first rider.

- physics -
**drwls**, Sunday, October 7, 2007 at 11:16pmFirst convert the speeds to convenient units of distance per second. I choose ft/s. You can convert to other distances later.

V1 = 30 mph = 44.00 ft/s

V2 = 40 mph = 58.67 ft/s

The distance of the first biker from the viewer is

X1 = 30 t

and the distance if the second biker (after 10 sec) is

X2 = 40 (t-10)

(t will be in seconds, measured from when the first biker goes by)

Set X1 = X2 and solve for t, and then solve for either X.

44 t = 58.67 (t-10) = 58.67 t - 586.7

14.67 t = 586.7

t = 40.0 s

Distance = 40 V1 = 1760 ft = 1/3 mile

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