Question

What are the magnitude δv and angle θ of the change in velocity if the initial velocity is 30.0 m/s south and the final velocity is 75.0 m/s west? Let the angle θ be measured relative to the east direction (e.G., θ=90∘ means δv⃗ points north, θ=180∘ means δv⃗ points west, etc.)?

Answer 1

Answer:

Magnitude of δv = 80.78 m/s

Angle , θ = $158.2^0$

Explanation:

   Here east direction is zero degree, that is zero degree is along positive X direction, north is along + Y direction, West is along - X direction and South is along -Y direction.

    In this case initial velocity = 30.0 m/s south = -30 j

                         Final velocity = 75.0 m/s west  = - 75 i

    We know that change in velocity =  Initial velocity - Final velocity

                                       So            δv = - 75 i - (-30 j)

                                                             =  - 75 i + 30 j m/s

   Magnitude of δv = $\sqrt{(-75)^2+30^2} = 80.78 m/s$

   Angle , θ = $tan^{-1}(30/(-75)) = 158.2^0$

Answer 2

The magnitude of the velocity is 80.78m/s

Since tan is negative in the third quadrant, hence $\theta = 180+21.8=201.8^0$

If the initial velocity moves towards the south, the velocity will be 30.0m/s j

If the final velocity is 75.0 m/s west, hence the direction will be -75m/s i

The magnitude of the velocity will be expressed as;

$v=\sqrt{v_x^2+v_y^2}\\v=\sqrt{(-75i)^2+(30j)^2}\\v=\sqrt{5625+900}\\ v=\sqrt{6525}\\v=80.78m/s$

Hence the magnitude of the velocity is 80.78m/s

Get the angle between the velocities

$\theta = tan^{-1}\frac{y}{x}\\ \theta = tan^{-1}\frac{30}{-75}\\\theta = tan^{-1}(-0.4)\\\theta = - 21.8^0$

Since tan is negative in the third quadrant, hence $\theta = 180+21.8=201.8^0$

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