Question

Two cars are moving at 60 mph in the same direction and in the same lane. The cars are separated by one car length (20 ft) for each 10 mph. The coefficient of friction (skidding) between the tires and the roadway is 0.6. The reaction time is assumed to be 0.5 sec. a) If the lead car hits a parked truck, what is the speed of the second car when it hits the first (stationary) car? b) If the lead car hits a parked truck and comes to an abrupt halt, at what speed does the rule of one car length of spacing for every 10 mph of speed become unsafe? c) At 60 mph, what should the rule actually be?

Answer 1

a) for 60mph = 6 x 20 = 120 ft is distance between them

speed of car = 60 x 5280ft / 3600sec = 88 ft/s

for the first 0.5 sec it will go with same speed ( reaction time)

d1 =0.5 x 88 = 44 ft

after brakes are applied,

a = -uk.g = 0.6 x 32.17 = - 19.30 ft/s

now distance left = 120 - 44 = 76 ft

using v^2 - u^2 = 2ad

v^2 - 88^2 = 2 x -19.30 x 76

v = 69.36 ft/s

b) suppose speed is u.

10mph =10 x 5280 / 3600 = 14.67 ft/sec for 20 ft

distance between cars. = (u/14.67) x 20 ft = 1.36u

reaction time distacne = 0.5u

distance left = 1.36 u - 0.5u = 0.86u

Using v^2 - u^2 = 2ad

0 - u^2 = 2 x -19.30 x 0.86u

u = 33.34 ft/s = (33.34 x 3600 / 5280) = 22.73 mph

c) suppose rule is y feet for every 10mph Or 14.67 ft/s

distance = 6y

speed of car = 60 x 5280ft / 3600sec = 88 ft/s

for the first 0.5 sec it will go with same speed ( reaction time)

d1 =0.5 x 88 = 44 ft

after brakes are applied,

a = -uk.g = 0.6 x 32.17 = - 19.30 ft/s

now distance left = 6y - 44

using v^2 - u^2 = 2ad

0^2 - 88^2 = 2 x -19.30 x (6y - 44)

6y - 44 =200.62

y = 40.77 ft