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Gnesinka [82]
3 years ago
12

A motorcycle is stopped at a traffic light. When the light turns green, the motorcycle accelerates to a speed of 91 km/h over a

distance of 47 m.
(a) What was the average acceleration of the motorcycle over this distance?

? m/s2

(b) Assuming the motorcycle maintained a constant acceleration, how far is it from the traffic light after 3.3 s?

? m
Physics
1 answer:
zimovet [89]3 years ago
6 0

Given :

Initial speed , u = 0 m/s .

Final speed , v = 91 km/h = 25.28 m/s .

To Find :

a) Average acceleration .

b ) Assuming the motorcycle maintained a constant acceleration, how far is it from the traffic light after 3.3 s .

Solution :

a )

We know ,by equation of motion :

v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{25.28^2-0^2}{2\times 47}\ m/s^2\\\\a=6.8\ m/s^2

b)

Also , by equation of motion :

s=ut+\dfrac{at^2}{2}\\\\s=0+\dfrac{6.8\times (3.3)^2}{2}\ m\\\\s=37.02\ m

Hence , this is the required solution .

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We apply this expression to our case

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The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical

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fiate that in the exercise we must take a reference system and measure everything with respect to this system.

          I = I₁ cos² θ'

       

we substitute

         I = (I₀ cos² tea) cos² (θ - 90)

        cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ

         I = Io cos² θ sin² θ

        1= cos²θ+ sin²θ

       sin²θ = 1 - cos²θ

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whereby

            cos θ - 2 cos³ θ = 0

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The zeros of this function are in

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