A car is traveling at the posted speed limit of 6.70 m/s (15 miles/h) in a school zone. The car passing a school bus when a child darts out from in front of the bus. The car is 10.0 m from the child when the driver of the car applies the brakes. The car decelerates at a constant rate of 4.00 m/s2 until coming to a halt. Calculate the
(a) distance of the car travels while decelerating and
(b) distance required to stop if the car is initially traveling at 13.4 m/s (30.0 miles/h).

5.6m and 22.4m

Explanation: from the 3rd equation of motion

v²=u²+2as

v²-u²=2as  where : v=0 u=6.7m/s a=4m/s² s=?

0-6.7²=2×4×s

44.89=8s

s=44.89÷8

s=5.6m

for the second part, we shall still make use of the same equation.

v²-u²=2as         where: a=4m/s² u=13.4m/s v=0 s=?

0-13.4²=2×4×s

179.56=8s

s=179.56/8

22.4m

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.

come closer and become far apart and become steam/gas