Question #a0eaf

Jan 31, 2018

Magnitude of Displacement$= 20 \sqrt{2} m$ = $\approx 28.284 m$
and
Average velocity =
${v}_{a v} \approx 0.14142 m {s}^{-} 1$

Explanation:

A girl moves along the boundary of a square field of side 20m in 80 secs.

Each side is 20m long.
So perimeter/ boundary is $4 \times 20 = 80 m$
That means she is moving $\frac{80}{80} \frac{m}{\sec}$ = $1 m$ in $1 \sec$

After 200 seconds she will have covered 200m.

The number of rounds covered along the complete boundary will be 200/ 80 = 2.5 rounds.

That means she is at the diagonally opposite point from where she started after 200 seconds.

So, the displacement will be given as distance between the initial position and final position, i.e diagonal of the square field. It will be hypotenuse of the right triangle with side = $20 m$

Apply Pythagoras theorem,

$\therefore$ Magnitude of Displacement$= 20 \sqrt{2} m$ = $\approx 28.284 m$

Average velocity = Displacement/ total time taken

${v}_{a v} = \frac{20 \sqrt{2}}{200} \frac{m}{s}$ = $\frac{\sqrt{2}}{10} \approx 0.14142 m {s}^{-} 1$

Jan 31, 2018

Magnitude of displacement

$\vec{d} = \sqrt{{20}^{2} + {20}^{2}} = \sqrt{800} = 20 \sqrt{2} \text{ m}$

Magnitude of average velocity

${\vec{v}}_{\text{avg"=sqrt2/10" m/s}}$

Explanation:

Assuming that the girl is moving at a constant speed for the 200 seconds, she will get around the field $\frac{200}{80} = 2.5$ times.

The displacement (notated by $\vec{d}$) is the distance of a straight line from the starting point to the end point. It is a vector, meaning that it has a magnitude and a direction. It is different from the distance travelled, which in this case is 2.5 times around the field ($200 \text{ m}$).

If you look at my crappy diagram below, you can see that the girl starts at the bottom left corner and ends up at the top right X. The magnitude of the displacement is the distance of the straight line from start to end point and can be solved using the Pythagorean Theorem. So, the displacement is

$\vec{d} = \sqrt{{20}^{2} + {20}^{2}} = \sqrt{800} = 20 \sqrt{2} \text{ m}$

Now, the average velocity (${\vec{v}}_{\text{avg}}$) is given by the displacement divided by the total time taken. It is also a vector, which means it has a magnitude and a direction.

${\vec{v}}_{\text{avg"=vecd/"time taken"=(20sqrt2)/200=sqrt2/10" m/s}}$

If we designated up the page as north, we could say that the average velocity and displacement both have directions of north-east.