A man drove his car north west for a distance of 10km,then east for 40km and then south for 60km.Calculate the overall displacement and direction of the car from the starting point.
the final location is <-10/√2,10/√2> + <40,0> + <0,-60>
So, evaluate the (x,y) value and then
displacement is √(x^2+y^2)
and the angle θ has tanθ = y/x
Well, it sounds like this man took quite the scenic route in his car! Let's calculate the overall displacement and direction for him, shall we?
First, let's break down his journey into the north, west, east, and south components.
The north-west component is a combination of the north and west directions, forming a right angle, so we can calculate its magnitude using the Pythagorean theorem: sqrt(10^2 + 0^2) = 10 km.
Next, the east direction doesn't affect the north-west component, so the overall displacement stays the same at 10 km.
Finally, going south adds a south component of 60 km.
Now, to find the overall displacement, we add up these components. The north-west and east components cancel each other out, and the remaining south component gives us a displacement of 60 km.
As for the direction, well, if you imagine standing at the starting point and pointing towards the final location, you'll be pointing directly south. So, the overall direction of the car from the starting point is "please proceed in a southerly fashion."
Hope that clears up the car's journey for you!
To calculate the overall displacement of the car, we need to find the net direction and distance traveled.
Step 1: Convert the directions to a common reference frame. Consider north as the positive y-axis and east as the positive x-axis.
Step 2: Calculate the north-south displacement. Since the car traveled south for 60km and north-west for 10km, the net displacement in the north-south direction is 60km - 10km = 50km south.
Step 3: Calculate the east-west displacement. The car traveled east for 40km, and since it drove northwest, we can use the Pythagorean theorem to find the north-west displacement. The northwest displacement can be calculated as the hypotenuse of a right-angled triangle with sides of 10km and 40km.
Using the Pythagorean theorem: northwest displacement = √(10km)^2 + (40km)^2 = √(100km^2 + 1600km^2) = √(1700km^2) ≈ 41.2km
Step 4: Determine the overall displacement. The overall displacement is the vector sum of the north-south and east-west displacements. The north-south displacement is 50km south, and the east-west displacement is 41.2km northwest.
Using the Pythagorean theorem again: overall displacement = √(50km)^2 + (41.2km)^2 = √(2500km^2 + 1694.4km^2) = √(4194.4km^2) ≈ 64.7km
Step 5: Determine the direction of the overall displacement. Use trigonometry to find the angle between the overall displacement vector and the positive x-axis.
Let θ be the angle between the overall displacement vector and the positive x-axis.
tan(θ) = (opposite side) / (adjacent side) = (50km / 41.2km) ≈ 1.21
θ ≈ 50.3°
The overall displacement of the car from the starting point is approximately 64.7km towards 50.3° east of south.
To calculate the overall displacement and direction of the car, we can use vector addition.
First, let's assign directions to each distance traveled:
- North is positive in the y-axis direction.
- East is positive in the x-axis direction.
- South is negative in the y-axis direction.
- West is negative in the x-axis direction.
Now, let's break down each leg of the journey:
1. The car drives northwest for a distance of 10 km. At a 45-degree angle, this can be split into its north and west components using trigonometry.
- North component: 10 km * cos(45°) = 10 km * √2/2 = 10 * 0.707 = 7.07 km north
- West component: 10 km * sin(45°) = 10 km * √2/2 = 10 * 0.707 = 7.07 km west
2. The car then drives east for a distance of 40 km. This only affects the x-axis direction.
3. Finally, the car drives south for a distance of 60 km. This only affects the y-axis direction.
To calculate the overall displacement, we add up the north-south and east-west components:
- North-south displacement: 7.07 km north - 60 km south = -52.93 km south
- East-west displacement: 40 km east - 7.07 km west = 32.93 km east
To find the overall displacement, we use the Pythagorean theorem:
Overall displacement = √((North-south displacement)^2 + (East-west displacement)^2)
= √((-52.93 km)^2 + (32.93 km)^2)
= √(2801.05 km^2 + 1083.05 km^2)
≈ √3884.1 km^2
≈ 62.29 km
So, the overall displacement of the car from the starting point is approximately 62.29 km.
To find the direction, we can use trigonometry again:
Direction = atan2(North-south displacement, East-west displacement)
= atan2(-52.93, 32.93)
≈ -58.3° or 301.7°
Since angles in vector notation are measured counterclockwise from the positive x-axis, the direction will be 301.7° if measured clockwise from the positive y-axis.
Therefore, the overall displacement of the car from the starting point is approximately 62.29 km in a direction of 301.7° clockwise from the positive y-axis.
All angles are measured CW from +y-axis.
AD = AB+BC+CD = 10km[315o] + 40km[90o] + 60km[180o],
AD=(10*sin315+40*sin90+60*sin180)+(10*cos315+40*cos90+60*cos180)I
AD = 32.93 - 52.93i = 62.3km[31.9o] S. of E. = 62.3km[121.9o] CW.
= Displacement.