If you walked 29752 meters in 2.00 hours what would your average speed be in m/s. then, suppose you slow down to 3.00 m/s at the midpoint, but then pick up again and accelerate to the speed calculated before. it takes you 30.0 s to accelerate. find the magnitude of the average acceleration during this time interval

Explanation:

x = 3.00t^{2} – 2.00t + 3.00,

Distance of object at 2 second,

x (t=2) = 3(4) - 2(2) +3

x (t=2) = 12-4 +3

x (t=2) = 11 m

Distance of object at 3 second,

x (t=3) = 3(9) - 2(3) +3

x (t=2) = 27 - 6 + 3

x (t=2) = 24 m

a) the average speed between t = 2.00 s and t = 3.00 s,

Average speed = \frac{Total distance}{ Total time}

Average speed = \frac{x (t=2) + x (t=3)}{3}

Average speed = \frac{24+11}{3}

Average speed = \frac{35}{3}

Average speed = 11.66 \frac{m}{s}

b) the instantaneous speed at t = 2.00 s and t = 3.00 s,

Instantaneous speed = \frac{dx}{dt}

Instantaneous speed(v) = 6t - 2\left \{ {{t=2} \atop {t=3}} \right.

Instantaneous speed,v(t=2 to t=3) = 18-2-12+2

Instantaneous speed, v = 6 \frac{m}{s}

c) the average acceleration between t = 2.00 s and t = 3.00 s

average acceleration = \frac{average velocity}{time}

average acceleration =  \frac{11.66}{3-2}

average acceleration = 11.66 \frac{m}{s^{2} }

d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s

instantaneous acceleration = \frac{dv}{dt}

instantaneous acceleration =6

instantaneous acceleration = 6 \frac{m}{s^{2} }

e) for x =0

0 = 3.00t^{2} – 2.00t + 3.00

a = 3, b=-2, c=3

t= \frac{-b \pm \sqrt{b^{2} - 4ac} }{2a}

t= \frac{2 \pm \sqrt{4 - 36} }{6}

t= \frac{2 \pm \sqrt{-32} }{6}

general solution of this equation gives imaginary value. Hence, the given object is not at rest.

Explanation:

Credit goes to @Branta

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.