The location of a car at times t1 and t2 is given by the position vectors
r1= 15km due east + 20km due north (time t1) and
r2= 13km due west + 25km due north (time t2)
The units of distance are kilometers
a. What is the magnitude l^rl of the displacement?
b. What angle does the displacement ^r vector make with respect to the x-axis?
c. Draw a graph of the given situation.
a. r1 = (15 + i20)km.
r2 = (-13 + i25)km.
r = r2 - r1\,
r = (-13 + i25) - (15 + i20),
r = -13 + i25 - 15 - i20,
Combine like-terms:
r = -28 + i5,
r = sqrt((-28)^2 + 5^2) =28.4km.
b. r = -28 + i5,
tanA = 5 / -28 = 0.17857,
A = -10.1 deg.,CW.
A = -10.1 + 180 = 169.9 deg.,CCW. Q2.
To find the displacement vector, we need to subtract the initial position vector (r1) from the final position vector (r2).
a. Displacement vector (Δr) = r2 - r1
= (13km west + 25km north) - (15km east + 20km north)
= (-2km west + 5km north)
To find the magnitude of the displacement (l^r1l), we can use the Pythagorean theorem.
l^r1l = √((-2km)^2 + (5km)^2)
= √(4km^2 + 25km^2)
= √(29km^2)
≈ 5.39 km
b. To find the angle θ that the displacement vector makes with respect to the x-axis, we can use trigonometry.
tan(θ) = opposite/adjacent
= 5km north / 2km west
= 2.5
θ = arctan(2.5)
≈ 68.2°
So, the displacement vector has a magnitude of approximately 5.39 km and makes an angle of approximately 68.2° with respect to the x-axis.
c. To draw a graph of the situation, we can use a coordinate system with the x-axis representing east and west, and the y-axis representing north and south.
Start by placing the initial position vector (r1) at the origin (0,0). From there, move 15km due east (to the right) and then 20km due north (upwards) to point A. Next, move 13km due west (to the left) from point A, and then 25km due north (upwards) to point B. The displacement vector (Δr) will start at point A (final position) and end at point B (initial position).