A car accelerates from rest at a constant acceleration of 25.0 m/s^2. At some point, it then turns off its engine, letting the car decelerate slowly from the force of friction at a constant deceleration of 3 m/s^2 until it is at rest again. The total speed the car moves in this time is 200 meters. What is the minimum time needed for the car to move 200 meters given that it both starts and ends at rest?

t = 9.14 s

Explanation:

We first analyze the accelerating motion by applying first equation of motion:

Vf₁ = Vi₁ + a₁t₁

where,

Vf₁ = Final Speed of Car before turning off engine

Vi₁ = Initial Speed of Car = 0 m/s

a₁ = acceleration of car = 25 m/s²

t₁ = time taken in accelerating motion

Therefore,

Vf₁ = 25t₁   equation (1)

Now, we apply second equation of motion:

s₁ = Vi₁ t₁ + (1/2)a₁t₁²

where,

s₁ = distance covered during accelerating motion

Therefore,

s₁ = (0)t₁ + (1/2)(25)t₁²

s₁ = 12.5 t₁²   equation (2)

Now, we analyze the decelerating motion by applying first equation of motion:

Vf₂ = Vi₂ + a₂t₂

where,

Vf₂ = Final Speed of Car = 0 m/s

Vi₂ = Initial Speed of Car after turning off engine

a₂ = deceleration of car = - 3 m/s²

t₂ = time taken in decelerating motion

Therefore,

Vi₂ = 3t₂   equation (3)

Now, we apply second equation of motion:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

s₂ = distance covered during decelerating motion

Therefore,

s₂ = (Vi₂)t₂ + (1/2)(-3)t₂²

s₂ = Vi₂ t₂ - 1.5 t₂²  

using equation (3):

s₂ = 3 t₂² - 1.5 t₂²

s₂ = 1.5 t₂²   equation (4)

Now, we know that the Final Velocity of accelerating motion (Vf₁) is equal to the initial velocity of decelerating motion (Vi₂):

Vf₁ = Vi₂

using equation (1) and equation (3):

25 t₁ = 3 t₂

t₁ = 0.12 t₂   equation (5)

Also, we know that sum of the distances is 200 m:

s₁ + s₂ = 200

using equation (2) and equation (4):

12.5 t₁² + 1.5 t₂² = 200

using equation (5):

12.5 (0.12 t₂²) + 1.5 t₂² = 200

3 t₂² = 200

t₂² = 200/3

t₂ = 8.16 s

substitute this in equation (5):

t₁ = 0.12(8.16 s)

t₁ = 0.97 s

Hence, the minimum time required for this motion is:

t = t₁ + t₂ = 0.97 s + 8.16 s

t = 9.14 s

positive reinforcement? ?

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.