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**Homework Statement**

The minimum safe distance between vehicles on a highway is the distance a vehicle can travel in 2.0s at a constant speed. assume that a 1.2x10^3 kg car is traveling 72km/h

a) if the car is at the required safe distance behind the truck, what is the separation distance?

b) If the average net braking force exerted by the car is 6.4x10^3 N [N], how long would it take the car to stop?

c) Determine whether a collision would occur. Assume that the driver's reaction time is an excellent 0.09s.

**The attempt at a solution**

3a) d = (vi+vf/2) t

= (1200+0/2) (2)

= 1200 m

b) Fnet = m*a

-6.4x10^3 = (1.2x10^2)a

a = -5.3 m/s^2

a = vf-vi/t

-5.3 = 0-1200/t

t = 226.4s

c) d = vit + 1/2at^2

d = (1200)(0.09) + 1/2 (-5.33)(0.09)^2

d = 108m

Which is the distance traveled in reaction

d = vit + 1/2at^2

= (1200)(225) + 1/2(-5.33)(225)^2

= 2.7x10^5 - 134915

= 135084m m

d = 13504 + 108

= 125192 m

so a collision would occur

Is this correct?