Posted by Amelia on Thursday, October 9, 2008 at 7:19pm.
Write equations for position vs time for both car and bike.
Solve for the time that the car passes the bike.
In other words, solve
X(car) = X(bike)
That will answer part (a).
For part (b), differentiate
X(bike) - X(car) vs. time and find out where the derivative is zero. That will be the time of maximum separation.
Show your work if you need additional help
First make everything feet and seconds
1 mi = 5280 ft
1 h = 3600 s
car
50 mi/h *5280 ft/mi *1h/3600s
= 73.3 ft/s
8.8 mi/hs *5280 ft/mi *1h/3600s =12.9ft/s^2
bike
20 mi/hr --> 29.3 ft/s
14.5 mi/hs --> 21.3 ft/s^2
Now as long as the bike is accelerating the bike will be getting further ahead. However once the car finally reaches 20 mph, the car will start to catch up.
So question (b) is easy. When does the car reach 20 mph or 29.3 ft/s ?
29.3 = 12.9 t
t = 2.27 s for the car to reach max bike speed, from then on it will be catching up.
How far did the car go before it started to catch up?
d = (1/2) (12.9) (2.27)^2=
33.2 ft (answer part b)
now when does the car finally pass the bike?
How long does the car accelerate?
73.3 = 12.9 t
t =5.68 s
during acceleration the car goes
d = .5 (12.9)(5.68^2) = 208 ft
from then on the car goes at 73.3 ft/s
How long does the bike accelerate?
29.3 = 21.3 t
t = 1.38 s
d = .5 (21.3)(1.38^2) = 20.1 ft
from then on the bike goes at 29.3 ft/s
Now probably the car passes the bike while the car is still accelerating so try that possibility first
d car = .5*12.9 * t^2
d bike = 20.1 + 29.3 (t-1.38)
6.45 t^2 = -20.3 +29.3 t
6.45 t^2 -29.3 t +20.3 = 0
t = 3.69 or .863
at .863 s the bike is ahead of the car and gaining, so that does not work
However at 3.69 s the car has gone
d = (1/2) (12.9) 3.69^2 = 87.8 ft
and the bike has gone
d = 20.1 + 29.3 (3.69-1.38) = 87.8 ft
so the bike was ahead for 3.69 s (part a)