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August 30, 2014

Posted by **Lisa** on Wednesday, April 14, 2010 at 1:19am.

(a) For how long is the bicycle ahead of the car?

(b) By what maximum distance does the bicycle lead the car?

- Physics -
**drwls**, Wednesday, April 14, 2010 at 5:51ama) Write equations for distance from the light vs time for bike and car. Call them X1(t) and X2(t)

Set X1 = X2 and solve for t. These are piecewise continous functions, so you cannot solve single equations.

If t is in seconds and X in feet,

X1(car) = 5.94 t^2

Car acceleration lasts for 6.17 seconds

X2(bike) = 9.533 t^2

Bike acceleration lasts for 1.538 s.

After 1.538s, the bike is ahead by X1-X2 = 22.55 - 14.05 = 8.5 feet. The car is traveling at 18.2 ft/s, which is slower than the bike's 29.3 ft/s but the car still accelerating. Let t' be the time measured after 1.538 seconds, when the bike reaches constant velocity.

X1 = 14.05 + 18.2 t' + 5.94 t'^2

X2 = 22.55 + 29.3 t'

You can now set up a quadratic equation for when X1 = X2

You can also differentiate X2 - X1 and set the derivative equal to zero to get the maximum separation time and distance.

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