PHYSICS 1 ..
posted by JOAN .
as she pick up her riders a bus driver traverses four successive displacements represented by the expression (-6.30 b)i-(4.00 b cos 40)i-(4.00 sin 40)j+(3.00 b cos 50)i-(3.00 b sin 50)j-(5.00 b)j here b represents one city block,a convenient unit of distance of uniform size; i is east; and j is north. the displacements at 40 degree and 50 degree represent travel on roadways in the city that are at these angles to the main east-west and north-south streets.(a)draw a map of the successive displacements(b)what total distance did she travel? (c)compute the magnitude and direction of her total displacements.
add up all your x (east) distances. I assume that they are all in b units and
-(4.00 sin 40)j
is a typo:
- 6.3 - 4(.766) + 3(.643) = -7.44 b east
(or 7.44 b locks west)
Add up your y (north) distances:
- 4(.643) - 3(.766) - 5 = -9.87 b North
(or 9.87 blocks south)
6.3+4+4+3+3+5 = 25.3
7.44 west and 9.87 south
magnitude^2 = 7.44^2+ 9.87^2
magnitude = 12.4 b
tan angle below -x axis = 9.87/7.44
angle = 53 degrees below -x axis
West 53 degrees South
217 degrees true on a compass
The distance is 18.3. 6.3+4+3+5
Total distance is 6.3+4+3+5= 18.3 blocks
Magnitude is 12.4b
tan^-1(-9.9/-7.4)= 53 but since it is in third quadrant, add 180 degrees to get a final angle of 233 degrees.