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inessss [21]
3 years ago
7

A particle moves along the x axis so that when t is greater than zero, it's position function is given by cos square root of t.

what is the velocity of the particle at the first instance when the particle is at the origin?
Physics
1 answer:
faust18 [17]3 years ago
4 0
The position is a cosine function, so the particle is at the origin
whenever the cosine is zero.
 
The first point where the cosine is zero occurs when the angle is  π/2 .
That happens when √t = π/2 , so t = π² / 4 is the point where we need
the particle's speed.

Speed is the first derivative of the position.
The derivative with respect to 't' of cos(√t) is [ -1 / (2√t) sin(√t) ] . (chain derivative.)

The speed when [ t = π² / 4 ] is . . .

-1 / 2√(π² / 4) times sin(√(π² / 4)) = -(1 / π) times sin(π/2) = -(1/π) times (1) = -(1/π) .

The first time when the particle is at the origin, it's moving backwards,
into [ -x ] territory, and its speed is (1/π) = about 0.3183... . (rounded)

There you have its speed and direction, so you have its velocity.
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The relationship of the speed of light

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The Planck equation is

          E = h f

Let's start the transformations

     c = f λ = f / ν        

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     T = h c ν  / K =( h c / K) ν

Let's replace the constants

     h = 6.63 10⁻³⁴ J s

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     v = 1 cm-1 (100 cm / 1 m) = 10² m-1

   

     T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 1 10²

     A = h c / K = 1,441 10⁻²

     T =  1.44K

     ν = 103 cm⁻¹ = 103 10² m

     T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 103 10²

     T = 148K

1 Rydberg = 1.097 10 7 m

As we saw at the beginning the λ=1 / v

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     T = 1,441 10⁻²  1 / 1,097 10⁷

     T = 1.3 10⁻⁹ K

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    T = E/K

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4 years ago
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Explanation:

The acceleration is defined as the rate of change of velocity.

Here, initial velocity, u = 3/1 = 3 cm/s

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