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University Physics: Mechanics
Ch4. TWO- AND THREE-DIMENSIONAL MOTION
Lecture 4
Dr.-Ing. Erwin Sitompulhttp://zitompul.wordpress.com
2013
4/2Erwin Sitompul University Physics: Mechanics
Solution of Homework 3: The Beetles
Starting point
1st run, 0.5 m
2nd run, 0.8 m
1st r
un, 1
.6 m
2nd run, ?
New location
30
40
4/3Erwin Sitompul University Physics: Mechanics
Thus, the second run of the green beetle corresponds to the vector
Solution of Homework 3: The Beetles
New location
Starting point
A
B
CD
30
40
ˆ ˆ ˆ0.5i (0.8cos30 i 0.8sin 30 j)A B
ˆ ˆ1.19i 0.4 j
ˆ ˆ1.6cos50 i 1.6sin 50 jC
ˆ ˆ1.03i 1.23j
→
→
→
→
A B C D
D A B C
ˆ ˆ ˆ ˆ(1.19i 0.4 j) (1.03i 1.23j) ˆ ˆ0.16i 0.83j
ˆ ˆ0.16i 0.83j m.D ��������������
4/4Erwin Sitompul University Physics: Mechanics
Solution of Homework 3: The Beetles
(a) The magnitude of the second run?
(b) The direction of the second run?
ˆ ˆ0.16i 0.83j mD
2 2(0.16) ( 0.83)D
0.85 m
1 0.83tan
0.16
79.09
N
E
S
W
D→
79.09
0.85 m
The direction of the second run is 79.09° south of due east or 10.91° east of due south.
4/5Erwin Sitompul University Physics: Mechanics
Moving in Two and Three Dimensions In this chapter we extends the material of the preceding
chapters to two and three dimensions. Position, velocity, and acceleration are again used, but they
are now a little more complex because of the extra dimensions.
4/6Erwin Sitompul University Physics: Mechanics
Position and Displacement One general way of locating a particle is with a position
vector r,ˆ ˆ ˆi j kr x y z
The coefficients x, z, and y give the particle’s location along the coordinate axes and relative to the origin.
The following figure shows a particle with position vectorˆ ˆ ˆ( 3 m)i (2 m) j (5 m)kr
In rectangular coordinates, the position is given by (–3 m, 2 m, 5 m).
→
4/7Erwin Sitompul University Physics: Mechanics
Position and Displacement As a particle moves, its position vector changes in a way that
the vector always extends from the origin to the particle. If the position vector changes from r1 to r2, then the particle’s
displacement delta is:
2 1r r r
2 2 2 1 1 1ˆ ˆ ˆ ˆ ˆ ˆ( i j k) ( i j k)r x y z x y z
2 1 2 1 2 1ˆ ˆ ˆ( )i ( )j ( )kr x x y y z z
ˆ ˆ ˆi j kr x y z
→ →
4/8Erwin Sitompul University Physics: Mechanics
Average Velocity and Instantaneous Velocity If a particle moves through a displacement Δr in a time
interval Δt, then its average velocity vavg is:
displacement
average velocitytime interval
avg
rv
t
The equation above can be rewritten in vector components as:
avg
ˆ ˆ ˆi j kx y zv
t
ˆ ˆ ˆi j kx y z
t t t
→
→
4/9Erwin Sitompul University Physics: Mechanics
The particle’s instantaneous velocity v is the velocity of the particle at some instant.
The direction of instantaneous velocity of a particle is always tangent to the particle’s path at the particle’s position.
Average Velocity and Instantaneous Velocity
drv
dt
→
4/10Erwin Sitompul University Physics: Mechanics
Average Velocity and Instantaneous Velocity Writing the last equation in unit-vector form:
This equation can be simplified by rewriting it as:ˆ ˆ ˆi j kx y zv v v v
where the scalar components of v are:
,x
dxv
dt ,y
dyv
dt z
dzv
dt
The next figure shows a velocity vector v and its scalar x and y components. Note that v is tangent to the particle’s path at the particle’s position.
ˆ ˆ ˆ( i j k)d
v x y zdt
ˆ ˆ ˆi j k
dx dy dz
dt dt dt
→
→
→
4/11Erwin Sitompul University Physics: Mechanics
The figure below shows a circular path taken by a particle. If the instantaneous velocity of the particle at a certain time is v = 2i – 2j m/s, through which quadrant is the particle currently moving when it is traveling(a) clockwise(b) counterclockwise
(a) clockwise (b) counterclockwise
Firstquadrant
Thirdquadrant
vy
vx
–2
2
ˆ ˆ2 2 m sv i j
Average Velocity and Instantaneous Velocity
^ ^→
4/12Erwin Sitompul University Physics: Mechanics
When a particle’s velocity changes from v1 to v2 in a time interval Δt, its average acceleration aavg during Δt is:
change in velocityaverage
acceleration time interval
2 1avg
v va
t
v
t
If we shrink Δt to zero, then aavg approaches the instantaneous acceleration a ; that is:
dva
dt
ˆ ˆ ˆ( i j k)x y z
dv v v
dt ˆ ˆ ˆi j kyx z
dvdv dv
dt dt dt
Average and Instantaneous Acceleration→ →
→
→
→
4/13Erwin Sitompul University Physics: Mechanics
We can rewrite the last equation asˆ ˆ ˆi j kx y za a a a
where the scalar components of a are:
,xx
dva
dt ,yy
dva
dt z
z
dva
dt
Acceleration of a particle does not have to point along the path of the particle
Average and Instantaneous Acceleration
→
4/14Erwin Sitompul University Physics: Mechanics
A particle with velocity v0 = –2i + 4j m/s at t = 0 undergoes a constant acceleration a of magnitude a = 3 m/s2 at an angle 130° from the positive direction of the x axis. What is the particle’s velocity v at t = 5 s?
Solution:0x x xv v a t 0y y yv v a t
0 2 m sxv 0 4 m syv 3 cos130xa 3 sin130ya
21.928 m s 22.298 m s
2 ( 1.928)(5)xv 11.64 m s
4 (2.298)(5)yv 15.49 m s
Thus, the particle’s velocity at t = 5 s is
ˆ ˆ11.64i 15.49j m s.v
Average and Instantaneous Acceleration^→
→
^→
At t = 5 s,
4/15Erwin Sitompul University Physics: Mechanics
Homework 4: The PlaneA plane flies 483 km west from city A to city B in 45 min and then 966 km south from city B to city C in 1.5 h. From the total trip of the plane, determine:(a) the magnitude of its displacement;(b) the direction of its displacement;(c) the magnitude of its average velocity;(d) the direction of its average velocity;(e) its average speed.
4/16Erwin Sitompul University Physics: Mechanics
Homework 4A: The TurtleA turtle starts moving from its original position with the speed 10 cm/s in the direction 25° north of due east for 1 minute. Afterwards, it continues to move south for 2 m in 8 s. From the total movement of the turtle, determine:(a) the magnitude of its displacement;(b) the direction of its displacement;(c) the magnitude of its average velocity;(d) the direction of its average velocity;(e) its average speed.
4/17Erwin Sitompul University Physics: Mechanics
Homework 4B: Shopping Trip1. A shopper at a supermarket follows the path indicated by vectors S, H, O,
and P in the figure. Given that the vectors have magnitudes S = 51 ft, H = 45 ft, O = 35 ft, and P = 13 ft. For all his movement, the shopper requires 8.5 min. Find:
→ → →
→
(a) the magnitude of his displacement;(b) the direction of his displacement;(c) the magnitude of his average velocity;(d) the direction of his average velocity;(e) his average speed.
2. Compute A + B + C and express the result in magnitude-angle notation (polar coordinate).
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