Question

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?​

Answer 1

The magnitude of displacement is 56.54 m

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

Step-by-step 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.

Answer 2

39.82 m, 22⁰ south west

Step-by-step 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

`(sine *\theta)/(35) = (sine*155)/(39.82)`

`{sine *\theta} = (35*sine155)/(39.82)`

`{sine *\theta} = (0.4226*35)/(39.82)`

`{sine *\theta}` = 0.3715

`\theta = sine^(-1) (0.3715)`

`\theta` = 21.81⁰ ≅22⁰

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