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A car is driven 160 km west and then 90 km southwest What is the displacement of the car from the point of origin magnitude and direction?
Wiki User
∙ 14y ago
*-----160km------*
|......................./
|...................../
x................../
|.............90km
|.............../
|............./
|.........../
|........./
|......./
|...../
|.../
*/
it's not to scale, but that's your model... now you use Pythagorean Theorem
x^2+160^2=90^2
x^2=-17500
x=the square root of -17500
Wiki User
∙ 14y ago
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A car is driven 12 km west then after a 90 degree turn drove 5 km south What is the magnitude of the displacement?
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A car is driven 215 km west and then 85 km southwest. What is the displacement of the car from the point of origin magnitude and direction draw a diagram.?
To solve this problem, I must explain the concept of vectors. Vectors merely consist of a magnitude and a direction. For this type of problem, the magnitude is the distance the car travels. Imagine arrows that are pointed in the direction of movement, and the same distance as the car moves. In this case, we will say that north is zero degrees. We know that since the car travels 215 km west, the first displacement is 215 west, and this is easy to visualize exactly where the car is. However, since the 85 km displacement is diagonal, it is more difficult to determine where exactly the vector goes. We must break this into components, in other words, two separate vectors. We must find out how far the car moves in the north-south direction, and how far it moves in an east-west direction. We do this using trigonometry. When we assumed that north is zero degrees, we determine that southwest corresponds to -135 degrees. So the calculations go as follows. For the east-west component, 85cos(-135)=-60.104 km. This means that the displacement from this vector is 60.104 km west. For the north-south component, 85sin(-135)= -60.104 km. This means that the displacement for this vector is 60.104 km south. We then add these two vectors to the 215km west. 215km + 60.104km = 275.104km. This means that the car has traveled a total of 275.104 km west. Since the car didn't travel south initally, we can just say that the car traveled 60.104 km south. To find out the straightline distance that this displacement is from the start, we use the pythagreaon theorem. The west and south displacements make up the legs of a right triangle. By adding the squares of these displacements, then taking the square root of the sum, we get 281.593 km from the start point. To get the angle of this displacement, use the inverse tangent function of the north-south component divided by the east-west component. We get 12.32 degrees. We must add this to the 90 degrees we get from the west component, so in the end, the vector can be defined as 281.593 km, -102.32 degrees. Hope this helps.
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