posted by Lisa on .
As soon as a traffic light turns green, a car speeds up from rest to 50.0 mi/h with a constant acceleration of 8.10 mi/h·s. In the adjoining bike lane, a cyclist speeds up from rest to 20.0 mi/h with a constant acceleration of 13.0 mi/h·s. Each vehicle maintains a constant velocity after reaching its cruising speed.
(a) For how long is the bicycle ahead of the car?
(b) By what maximum distance does the bicycle lead the car?
a) 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.