Tests reveal that a normal driver takes about 0.75s before he or she can react to a situation to avoid a collision. it takes about 3s for a driver having 0.1% alcohol in his system to do the same.

if such drivers are traveling on a straight road at 30mph (44 )ft/s and their cars can decelerate at 24ft/s^2 , determine the shortest stopping distance d for normal driver from the moment he or she see the pedestrians.

also, determine the shortest stopping distance for drunk driver from the moment he or she see the pedestrians.

NORMAL DRIVER: d = 73.3 ft

DRUNK DRIVER: d = 172.3

Explanation:

NORMAL DRIVER:

Distance covered in initial 0.75s = 0.75s *44 =  33ft

USING THE THIRD EQUATION OF MOTION

V^2-U^2 = 2as

0-(44)^2 = 2 (-24) s

s = 1936/48 =40.3 ft

d = 33 + 40.3 = 73.3 ft

DRUNK DRIVER:

Distance covered in initial 3s = 3s *44 =  132 ft

USING THE THIRD EQUATION OF MOTION

V^2-U^2 = 2as

0-(44)^2 = 2 (-24) s

s = 1936/48 =40.3 ft

d = 132 + 40.3 = 172.3 ft

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

ice to water 1 water to vapour 2 latent heat