A puck on the ice travels 20.0 m [25° W of N], gets deflected, and travels 30.0 m [35° N of W]. Determine where the puck will end up with re

Question

A puck on the ice travels 20.0 m [25° W of N], gets deflected, and travels 30.0 m [35° N of W]. Determine where the puck will end up with respect to its starting point, which is the puck’s total displacement, using the tail-to-tip with math (trigonometry) method. In the same hockey game, a player is positioned 35 m [40° W of S] of the net. He shoots the puck 25 m [E] to a teammate. What second displacement does the teammate shoot the puck with in order to make it to the net? Find the answer using the component method.

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Verity 4 years 2021-08-23T03:20:53+00:00 1 Answers 33 views 0

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  1. taiduc
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    2021-08-23T03:22:49+00:00
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    Given that,

    Ice travel = 20.0 m

    The direction is 25° W of N

    Deflect distance = 30.0 m

    The direction is 35° N of W

    (I). When ice travel west of north

    We need to calculate the displacement

    Using formula of displacement of component

    \vec{D}=d\cos\theta+d\sin\theta

    Put the value into the formula

    \vec{D}=20\cos25(j)+20\sin25(-i)

    \vec{D}=18j-8.44i

    (II). When ice deflect north of west

    We need to calculate the displacement

    Using formula of displacement of component

    \vec{D'}=d\cos\theta+d\sin\theta

    Put the value into the formula

    \vec{D'}=30\cos35(-i)+30\sin35(j)

    \vec{D'}=-24.3i+17.1j

    We need to calculate the magnitude puck’s total displacement

    Using formula for total displacement

    \vec{D''}=\vec{D}+\vec{D'}

    Put the value into the formula

    \vec{D''}=18j-8.44i-24.3i+17.1j

    \vec{D''}=-32.74i+35.1j

    |\vec{D''}|=\sqrt{(-32.74)^2+(35.1)^2}

    D''=47.9\ m

    Hence, The magnitude puck’s total displacement is 47.9 m.

    (ii). Given that,

    He shoot the puck 25 m in east.

    The displacement is

    \vec{D}=25i+0j

    A player is positioned 35 m at 40° West of south.

    We need to calculate the displacement

    Using formula of displacement

    \vec{D'}=d\cos\theta(-i)+d\sin\theta(-j)

    Put the value into the formula

    \vec{D'}=35\cos40(-i)+35\sin40(-j)

    \vec{D'}=-26.81i-22.49j

    We need to calculate the total displacement

    Using formula for total displacement

    \vec{D''}=\vec{D}+\vec{D''}

    Put the value into the formula

    \vec{D''}=25i+0j-26.81i-22.49j

    \vec{D''}=-1.81i-22.49j

    The magnitude of the total displacement

    |\vec{D''}|=\sqrt{(-1.81)^2+(-22.49)^2}

    D''=22.56\ m

    Hence, The magnitude of the total displacement is 22.56 m

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