Initial velocity vector vA has a magnitude of 3.00 meters per second and points 20.0o north of east, while final velocity vector vB has a ma

Question

Initial velocity vector vA has a magnitude of 3.00 meters per second and points 20.0o north of east, while final velocity vector vB has a magnitude of 6.00 meters per second and points 40.0o south of east. Find the magnitude and the direction of the change in velocity vector Δv (which is the vector subtraction of the two vectors: final velocity vector minus initial velocity vector).

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Thành Công 5 years 2021-09-02T20:02:53+00:00 1 Answers 14 views 0

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  1. Gerda
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    2021-09-02T20:04:16+00:00
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    Answer:

    5.2\ \text{m/s}

    70^{\circ} south of east

    Explanation:

    v_a = 3 m/s

    \theta_a = 20^{\circ} north of east

    v_b = 6 m/s

    \theta_b = 40^{\circ} south of east = 360-40=320^{\circ} north of east

    x and y component of v_a

    v_{ax}=v_a\cos \theta\\\Rightarrow v_{ax}=3\times \cos 20^{\circ}\\\Rightarrow v_{ax}=2.82\ \text{m/s}

    v_{ay}=v_a\sin\theta\\\Rightarrow v_{ay}=3\times \sin20^{\circ}\\\Rightarrow v_{ay}=1.03\ \text{m/s}

    x and y component of v_b

    v_{bx}=v_b\cos \theta\\\Rightarrow v_{bx}=6\times \cos 320^{\circ}\\\Rightarrow v_{bx}=4.6\ \text{m/s}

    v_{by}=v_b\sin\theta\\\Rightarrow v_{by}=6\times \sin320^{\circ}\\\Rightarrow v_{by}=-3.86\ \text{m/s}

    \Delta v=v_b-v_a\\\Rightarrow \Delta v=(4.6-2.82)\hat{i}+(-3.86-1.03)\hat{j}\\\Rightarrow \Delta v=1.78\hat[i}-4.89\hat{j}

    Magnitude

    |\Delta v|=\sqrt{(-4.89)^2+1.78^2}\\\Rightarrow \Delta v=5.2\ \text{m/s}

    Direction

    \theta=\tan{-1}|\dfrac{-4.89}{1.78}|\\\Rightarrow \theta=70^{\circ}

    The magnitude of the change in velocity vector is 5.2\ \text{m/s} and the direction is 70^{\circ} south of east.

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