A car travelled a distance of 5km due East, then 5km due North and then 5km due East. Calculate the total distance and displacement travelled by car also sketch the path.

Respuesta :

Answer:

Total displacement

The displacement of the car is 10 kilometers due East and 5 kilometers due North.

The magnitude of the displacement of the car is approximately 11.180 kilometers.

Total distance

The total distance of the car is 15 kilometers.

Explanation:

According to the statement the car shows the following path, whose displacement is represented by the following formula: (All distances are measured in kilometers)

[tex]\vec r = \vec r_{1} + \vec r_{2}+\vec r_{3}[/tex] (1)

Where:

[tex]\vec r_{1}[/tex] - First displacement to the east.

[tex]\vec r_{2}[/tex] - Displacement to the north.

[tex]\vec r_{3}[/tex] - Second displacement to the east.

If we know that [tex]\vec r_{1} = (5,0)[/tex], [tex]r_{2} = (0,5)[/tex] and [tex]r_{3} = (5,0)[/tex], then the displacement of the car is:

[tex]\vec r = (10, 5)[/tex]

The displacement of the car is 10 kilometers due East and 5 kilometers due North.

The magnitude of the displacement represents the distance of the car from point of departure in a straight line and is determined by the Pythagorean Theorem:

[tex]\|\vec r\| = \sqrt{10^{2}+5^{2}}[/tex]

[tex]\|\vec r\| \approx 11.180\,km[/tex]

The magnitude of the displacement of the car is approximately 11.180 kilometers.

The distance travelled of the car is the sum of magnitudes of the displacement of the car, each of them are calculated by Pyhtagorean Theorem:

[tex]d = \|\vec r_{1}\| + \|\vec r_{2}\| + \|\vec r_{3}\|[/tex] (2)

[tex]d = 5\,km + 5\,km + 5\,km[/tex]

[tex]d = 15\,km[/tex]

The total distance of the car is 15 kilometers.