At t = 0, a car registers at 30 miles/hr. Forty seconds later, the car’s velocity is now at 50 miles/hr. Assuming constant acceleration, what is the distance covered after 40s? Select one:
a. 0.442 mi
b. .33 mi
c. 0.11 mi
d. 10.33
e. None of the choices given.

Choice A) 0.44 miles

Work Shown:

1 hour = 60 minutes

1 minute = 60 seconds

1 hour = 60*60 = 3600 seconds

40 seconds = 40*(1/3600) = 40/3600 = 1/90 hours

The car travels 1/90 of an hour. Let t = 1/90.

The initial velocity is vi = 30 mph.

The final velocity is vf = 50 mph.

Apply one of the kinematics equations as given below.

x = distance traveled

x = 0.5*(vi+vf)*t

x = 0.5*(30+50)*(1/90)

x = 0.44444444444444 which is approximate

x = 0.44

The car traveled roughly 0.44 miles. This matches with choice A.

I'm assuming choice A is supposed to say 0.44 instead of 0.442; otherwise, I think your teacher made a typo by putting that 2 in there. Each of the other answer choices are accurate to 2 decimal places, so it would make sense that choice A is also accurate to 2 decimal places as well.

If 0.442 was intended by your teacher, and it's not a typo, then the answer would be E) None of the choices given.

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle  

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.