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

The leader in a bicycle race passes your viewing position at t=0 with a speed of 30 mph. Ten seconds later, the next racer goes by at 40 mph. Assuming their speeds are constant, where will the second cyclist catch up to the first? (Give this as a distance from your position)

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 11:14pm
First 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 11:16pm
First 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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