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1 A football player runs directly down the field for 35 m before turning to the right at
an angle of 25' from his original direction and running an additional 15 m before
getting tackled. What is the magnitude and direction of the runner's total
displacement?​

Respuesta :

Answer:

The magnitude of displacement is 56.54 m

The direction of the displacement is along the line joining the two vectors.

Explanation:

The resultant displacement is always the line joining the initial and final position of the vectors.

As in figure,

              the vector AB = 35 m

              the vector BC =  15 m

              the angle between AB and AC = 25'  (minutes)

              the resultant vector AC = ?

The resultant vector is given by the formula

             AC² = AB² + BC² + 2 AB BC Cos θ

Substituting the values in the equations,

             AC²  = 35² + 15² + 2 x 35 x 15 x Cos 25'

                      = 56.54

Therefore, the magnitude of displacement is 56.54 m

The direction of the displacement is along the line joining the two vectors.

Ver imagen jitendrasisodiyaj821
onyebuchinnaji

Answer:

39.82 m, 22⁰ south west

Explanation:

Step 1: make a sketch of the players displacement as shown in the image uploaded

Step 2: calculate the resultant displacement (R) from the image uploaded using cosine rule

R² = 35² + 15² -2(35*15)*cos(155)

R² = 1225 + 225 - 1050 *(-0.906)

R² = 1450 + 135.9

R² = 1585.9

R = √1585.9

R = 39.82 m

Step 3: calculate the players position(Ф) as shown in the image uploaded using sine rule

[tex]\frac{sine *\theta}{35} = \frac{sine*155}{39.82}[/tex]

[tex]{sine *\theta} = \frac{35*sine155}{39.82}[/tex]

[tex]{sine *\theta} = \frac{0.4226*35}{39.82}[/tex]

[tex]{sine *\theta}[/tex] = 0.3715

[tex]\theta = sine^{-1} (0.3715)[/tex]

[tex]\theta[/tex] = 21.81⁰ ≅22⁰

Therefore, the magnitude  and direction of the runner total displacement is 39.82m and 22⁰ south west respectively.

Ver imagen onyebuchinnaji

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