Posted by Anonymous on Saturday, September 7, 2013 at 2:55pm.
CAR:
V1 = 45mi/h * 1600m/mi * 1h/3600s=20 m/s
a=8.5mi/h/s*1600m/mi*1h/3600s=3.78m/s^2
a*t = 20 m/s
3.78t = 20
t = 5.29 s. to reach max velocity.
d1 = 0.5a*t^2
d1 = 0.5*3.78*5.29^2 = 52.9 m = Dist.@
which max. velocity is reached.
BIKE:
V2 = (15/45)*20 m/s = 6.67 m/s = max.
velocity.
a = (12.5/8.5)*3.78 = 5.56 m/s^2.
a*t = 6.67 m/s
5.56t = 6.67
t = 1.2 s to reach max velocity.
d2 = 0.5a*t^2
d2 = 0.5*5.56*1.2^2 = 4.0 m = Dist. @
which max. velocity is reached.
d1 = 0.5*3.78*1.2^2 = 2.72 m. = Dist.of
the car after 1.2 s.
D = d2-d1 = 4-2.72 = 1.28 m = Dist. the
by which the bike leads after 1.2 s.
When the car catches up:
a. d1 = d2 + 1.28 m.
0,5*a*t^2 = V2*t + 1.28
0.5*3.78*t^2 = 6.67*t+1.28
1.89t^2-6.67t-1.28 = 0
t = 3.71 s.(Use Quadratic Formula).
b. D = d2-d1 = 4-2.72 = 1.28 m.
a. t = (V-Vo)/a = (15-0)/12.5 = 1.2 s.
b. a=12.5mi/h/s*1600m/mi*1h/3600s=5.56m/s^2
= Acceleration of bike.
Db = 0.5*5.56*1.2^2 = 4 m. = Dist. of bike after 1.2 s.
a = (8.5/12.5) * 5.56 = 3.78 m/s^2 =
Acceleration of the car.
Dc = 0.5*3,78*1.2^2 = 2.72 m. = Dist. of the car after 1.2 s.
D = Db - Dc = 4 - 2.72 = 1.28 m