A car is driven east for a distance of 49 km, then north for 21 km, and then in a direction 30° east of north for 27 km. Determine (a) the magnitude (in km) of the car's total displacement from its starting point and (b) the angle (from east) of the car's total displacement measured from its starting direction.
30deg E of N = 60deg CCW.
X = hor = 49 + 27*cos60,
X=49 + 13.5=62.5km=hor comp. of vector.
Y = ver = 21 + 27*sin60,
Y = 21 + 23.38 = 44.38km.
a. D = sqrt(x^2 + y^2),
D = sqrt(3906.25+1969.58),
D = 76.7km.
b. tanA = Y/X = 44.38 / 62.5 = 0.7101.
A = 35.4 deg = Direction.
To determine the car's total displacement, we need to find the magnitude and direction of the vector sum of its individual displacements.
(a) To find the magnitude of the car's total displacement, we can use the Pythagorean theorem to calculate the hypotenuse of a right triangle formed by the car's eastward and northward displacements.
Using the given values:
Eastward displacement = 49 km
Northward displacement = 21 km
Magnitude of the total displacement = √(49^2 + 21^2)
= √(2401 + 441)
= √(2842)
≈ 53.33 km
Therefore, the magnitude of the car's total displacement is approximately 53.33 km.
(b) To find the angle of the car's total displacement measured from its starting direction, we can use trigonometry.
Using the given values:
Eastward displacement = 49 km
Northward displacement = 21 km
Displacement in a direction 30° east of north = 27 km
The total displacement vector forms a triangle with the eastward and northward displacements as adjacent sides. To find the angle, we need to find the angle opposite the eastward displacement.
Using the Law of Cosines:
cos(angle) = [(27^2 + 49^2) - 21^2] / [2 * 27 * 49]
angle = arccos([(27^2 + 49^2) - 21^2] / [2 * 27 * 49])
angle ≈ 39.82°
Therefore, the angle of the car's total displacement measured from its starting direction is approximately 39.82° east of east.
To determine the car's total displacement, we can use vector addition. We'll break down each leg of the journey into x and y components and then add them together.
Let's start with the first leg: driving east for 49 km. Since this is a purely eastward direction, the x-component of the displacement is 49 km (as there is no vertical component).
Next, let's consider the second leg: driving north for 21 km. Since this is a purely northward direction, the y-component of the displacement is 21 km (as there is no horizontal component).
Now, let's analyze the third leg: driving in a direction 30° east of north for 27 km. To determine the x and y components, we'll use trigonometry. The vertical component (y) is given by y = 27 km * sin(30°), which equals 13.5 km. The horizontal component (x) is given by x = 27 km * cos(30°), which equals 23.382 km (rounded to three decimal places).
Now, we can add up the x and y components to find the total displacement. The x-component sum is 49 km + 23.382 km = 72.382 km (rounded to three decimal places). The y-component sum is 21 km + 13.5 km = 34.5 km.
To find the magnitude of the total displacement, we can use the Pythagorean theorem. The magnitude (d) is given by d = sqrt(x^2 + y^2), where x is the horizontal component and y is the vertical component. Therefore, d = sqrt((72.382 km)^2 + (34.5 km)^2) = sqrt(5247.301 km^2 + 1190.25 km^2) = sqrt(6437.551 km^2) = 80.23 km (rounded to two decimal places).
So, the magnitude of the car's total displacement from its starting point is approximately 80.23 km.
To find the angle of the total displacement measured from its starting direction, we can use the inverse tangent function. The angle (θ) is given by θ = arctan(y/x), where y is the vertical component and x is the horizontal component. Therefore, θ = arctan(34.5 km/72.382 km) ≈ 26.8° (rounded to one decimal place).
So, the angle of the car's total displacement measured from its starting direction (east) is approximately 26.8°.