A meteorologist tracks the movement of a thunderstorm with Doppler radar. At 8:0
ID: 1531051 • Letter: A
Question
A meteorologist tracks the movement of a thunderstorm with Doppler radar. At 8:00 PM, the storm was 55 minortheast of her station. At 11:00 PM, the storm is at 75 mi north. What is the direction of the average velocity of the storm?
What is the magnitude of the average velocity of the storm?
A student bikes to school by traveling first dN = 0.800 miles north, then dW = 0.300 miles west, and finally dS = 0.200 miles south. Take the north direction as the positive y direction and east as positive x. The origin is still where the student starts biking. Let d N be the displacement vector corresponding to the first leg of the student's trip. Express d N in component form.
Explanation / Answer
(1) Find the position vectors
A = 55cos45 i + 55sin45 j
= 38.89 i + 38.89 j
B = 75 j
Displacement d = B - A
= -38.89 i + 36.10 j
direction
tan tita = (36.10/-38.89)
= -42.87degree
= -42.87 + 180 (to get into QIII)
= 137.13 degree from +x axis.
mag d = (38.89 + 36.10) mi = 75 mi
mag V = mag d / time
= 75 mi / 3 h = 25 mi/h
(2) Student bikes to school by traveling first dN = 0.800 miles north
Corresponding vector (V1)= 0 i + 0.8 j
student then bikes to school by traveling dW = 0.300 miles west
Corresponding Vector (V2) = -0.3 i + V1
= -0.3 i + (0 i + 0.8 j)
= -0.3 i + 0.8 j
student then bikes to school by traveling dS = 0.200 miles south
Corresponding Vector (V3) = -0.2 j + V2
= -0.2 j + ( -0.3 i + 0.8 j)
= -0.3 i + 0.6 j
dN = -0.3 i + 0.6 j