Question

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

Answer 1

Answer:

The magnitude of the displacement is 48.97 m and direction is $7.4^{\circ}$ to the right of his original position.

Explanation:

Given values

A distance ran by football player = 35 m

B distance ran additionally = 15 m

θ angle turned by the football player = 25°

Calculating magnitude:

Total displacement is “c” and calculated by

$C^{2}=\left(C_{x}^{2}+C_{y}^{2}\right)$

Where,

$C_{x}=A_{x}+B_{x}$

$C_{y}=B_{x}+B_{y}$

$A_{x}=0, A_{y}=35$

$B_{x}=B \sin \theta$

$\mathrm{B}_{\mathrm{y}}=\mathrm{B} \cos \theta$

Now, $C_{x}=A_{x}+B_{x}$

$C_{x}=0+15 \sin 25^{\circ}$

$C_{x}=15 \times 0.422$

$C_{x}=6.33 \mathrm{m}$ and

$C_{y}=A_{y}+B_{y}$

$C_{y}=35+15 \cos 25^{\circ}$

$C_{y}=35+25 \times 0.906$

$C_{x}=35+13.59$

$C_{y}=48.6 \mathrm{m}$

Total displacement is

$c=\sqrt{\left(c_{x}^{2}+c_{y}^{2}\right)}$

$C=\sqrt{6.33^{2}+48.6^{2}}$

$\mathrm{C}=\sqrt{40+2361.96}$

$C=\sqrt{2401.96}$

C = 49 m

Calculating Direction:

$\tan \Phi=\frac{C_{x}}{C_{y}}$

$\Phi=\tan ^{-1}\left(\frac{C_{x}}{C_{y}}\right)$

$\Phi=\tan ^{-1}\left(\frac{6.33}{48.6}\right)$

$\Phi=\tan ^{-1} 0.13$

$\Phi=7.4^{\circ} \text { to the right of down field }$

The magnitude of the displacement is 48.97 m and direction is $7.4^{\circ}$ to the right of his original position.