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

The magnitude of the displacement is 48.97 m and direction is
`7.4^(\circ)` to the right of his original position.

Step-by-step 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 * 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 * 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}=√(40+2361.96)`

`C=√(2401.96)`

C = 49 m

Calculating Direction:

`\tan \Phi=(C_(x))/(C_(y))`

`\Phi=\tan ^(-1)\left((C_(x))/(C_(y))\right)`

`\Phi=\tan ^(-1)\left((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.