# Question e7d7a

Jan 26, 2018

$= 500 m$ at an average velocity of $\frac{250 m}{\min}$ towards the NW

#### Explanation:

To answer this, one should be guided by the following facts that a displacement;

1. is a straight line drawn from its original position to the object's final position.
2. is a vector quantity; thus give attention to the direction of motion, but if the direction is one-dimensional motion; direction can be represented as positive or negative. Take note of the following:
$\text{upward (north)=positive; downward (south)=negative}$
$\text{left (west)=negative; right (east)=positive}$
3. is not always equal to the distance being traveled.

This case, the car has traveled $300 m \text{ North(N)}$, 400m" West(W);,# and that, when we connect the initial position to its final position, apparently is forming a right triangle; thus, the
he concept of the Pythagorean Theorem can be applied to find the resultant length of the travel or its displacement.

$\text{Pythagoream theorem}$ states that the square of the hypotenuse of a right triangle is equal to the squares of its legs; that is,

${c}^{2} = {a}^{2} + {b}^{2}$

The length of the line that connects the final and initial positions of the car can be computed as;

$c = \sqrt{{a}^{2} + {b}^{2}}$
where:

$c = \text{the hypotenuse=the resultant length}$
$a = \text{other leg=300m North}$
$b = \text{other leg=400m West}$

$c = \sqrt{{300}^{2} + {400}^{2}}$
$c = \sqrt{90 , 000 + 160 , 000}$
$c = \sqrt{250 , 000}$
$c = 500 m$

Therefore the displacement is $500 m$ at an average velocity of $\frac{250 m}{\min}$ towards the $\text{Northwest"(NW)} .$