Question

selasie2. A particle is displaced from the point (-8 m, -10 m) to the point (1 m, 1 m).Calculate the magnitude of its displacement vector. (1 point)A. O14.213 mB. O 202 mC. O 12.728 mD. O 11.402 mE. O 130 m3. A particle is displaced from the point (-4 m, -8 m) to the point (9 m, -1 m) in 30 secCalculate the magnitude of its average velocity. (1 point)A. O 0.409 m

Answer 1

A)14.213 m

Explanation

the distance between 2 points can be found by usign the formula:

$\begin{gathered} d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ where \\ P1(x_1,y_1) \\ P2(x_2,y_2) \end{gathered}$

This formula is derived from the Pythagorean theorem.

Step 1

given

$\begin{gathered} P1(-8,-10) \\ P2(1,1) \end{gathered}$

now, replace in the formula

$\begin{gathered} d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ d=\sqrt{(1-(-10))^2+(1-(-8))^2} \\ d=\sqrt{(11)^2+(9)^2} \\ d=\sqrt{121+81} \\ d=\sqrt{202} \\ d=14.213 \end{gathered}$

therefore, the answer is

A)14.213 m

I hope this helps you