Question

The site from which an airplane takes off is the origin. The X axis points east, the y axis points straight up. The position and velocity vectors of the plane at a later time are given by r=(1.21x103i+3.45x104;)m and v= (2 i-3.5j) m/s The magnitude, in meters, of the plane's displacement from the origin is:_________

a. 2.50 x104
b. 1.45 x 104
c. 3.45x104
d. 2.5x103
e. none of the above

Answer 1

The magnitude of the airplane's displacement from the origin is approximately 3.45 × 10 meters, which corresponds to option c.

Step-by-step explanation:

The magnitude of the plane's displacement from the origin can be found by using the Pythagorean theorem with the components of the position vector r. The position vector r given is (1.21 × 103i + 3.45 × 104j) meters. To find the magnitude, we calculate the square root of the sum of the squares of its components:

|r| = √((1.21 × 103)×2 + (3.45 × 104)×2)

When you work this out, you will find that the magnitude is approximately 3.45 × 104 meters, which matches option c.

Answer 2

d = 3.5*10^4 m

Step-by-step explanation:

In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:

`\vec{r}=(1.21*10^3\hat{i}+3.45*10^4\hat{j})m`

The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:

`d=√((x-x_o)^2+(y-y_o)^2)` (1)

where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):

`d=√((1.12*10^3-0)^2+(3.45*10^4-0)^2)=3.45*10^4m`

hence, the displacement of the airplane is 3.45*10^4 m