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September 30, 2014

September 30, 2014

Posted by **Anonymous** on Saturday, September 7, 2013 at 2:55pm.

(a) For what time interval is the bicycle ahead of the car?

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

- Physics -
**Henry**, Sunday, September 8, 2013 at 5:40pmCAR:

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

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