An electric vehicle starts from rest and accelerates at a rate of 2.0 m/s2 in a straight line until it reaches a speed of 21 m/s. The vehicle then slows at a constant rate of 1.1 m/s2 until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle move from start to stop?

total time = 29.59 sec

total distance = 310.70 m

Explanation:

given data

accelerates at a rate = 2.0 m/s²

speed = 21 m/s

slows at a constant rate = 1.1 m/s²

solution

first we take  equation of motion

v = u + at        1

21 = 0 + 2 × t

t = 10.5 sec

now in case of deceleration again we take here

0 = 21 - 1.1 × t

t = 19.09 sec

so that total time is

total time = 10.5 + 19.09

total time = 29.59 sec

and

in case of acceleration

v² = u² + 2as           2

21² = 0 + 2 × 2 × s

s = 110.25 m

and when it is in case of deceleration we take here

0 = 21² - 2 × 1.1 × s

s = 200.45 m

so that here total distance is

total distance = 110.25  + 200.45

total distance = 310.70 m

i think they are all correct

explanation:

they seem good

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(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.