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Unformatted text preview: MECHANICS Kinematics Chapter 2  Motion along a straight line
I. Position and displacement II. Velocity
III. Acceleration
IV. Motion in one dimension with constant acceleration
V. Free fall Particle: pointlike object that has a mass but infinitesimal size. I. Position and displacement
Position: Defined in terms of a frame of reference: x or y axis in 1D.
 The object’s position is its location with respect to the frame of reference.
PositionTime graph: shows the motion of
the particle (car). The smooth curve is a guess as to what happened between the data points. 1 I. Position and displacement
Displacement: Change from position x1 to x2
during a time interval. ∆x = x2x1 (2.1)  Vector quantity: Magnitude (absolute value) and direction (sign).
 Coordinate (position) ≠ Displacement x ≠ ∆x x x Coordinate system x1=x2 x2
x1 t
∆x >0 t
∆x = 0 Only the initial and final coordinates influence the displacement
different motions between x1and x2 give the same displacement. many Distance: length of a path followed by a particle.
 Scalar quantity
Displacement ≠ Distance
Example: round trip houseworkhouse distance traveled = 10 km
displacement = 0 Review:
 Vector quantities need both magnitude (size or numerical value) and
direction to completely describe them.
 We will use + and – signs to indicate vector directions.
 Scalar quantities are completely described by magnitude only. 2 II. Velocity Average velocity: Ratio of the displacement ∆x that occurs during a
particular time interval ∆t to that interval.
v avg = ∆x x 2 − x 1
=
∆t
t 2 − t1 Motion along xaxis (2.2) Vector quantity
indicates not just how fast an
object is moving but also in which direction it is
moving.
 SI Units: m/s
 Dimensions: Length/Time [L]/[T]  The slope of a straight line connecting 2 points on
an xversust plot is equal to the average velocity
during that time interval. Average speed: Total distance covered in a time interval. Savg = Total distance
∆t (2.3) Savg ≠ magnitude Vavg
Savg always >0 Scalar quantity
Same units as velocity
Example: A person drives 4 mi at 30 mi/h and 4 mi and 50 mi/h
average speed >,<,= 40 mi/h ?
<40 mi/h
t1= 4 mi/(30 mi/h)=0.13 h ; t2= 4 mi/(50 mi/h)=0.08 h Is the ttot= 0.213 h Savg= 8 mi/0.213h = 37.5mi/h 3 Instantaneous velocity: How fast a particle is moving at a given instant.
vx = lim ∆t → 0 ∆x dx
=
∆t dt (2.4)  Vector quantity  The limit of the average velocity as the time interval becomes infinitesimally
short, or as the time interval approaches zero.
 The instantaneous velocity indicates what is happening at every point of time.
 Can be positive, negative, or zero. x(t)  The instantaneous velocity is the slope of
the line tangent to the x vs. t curve
(green line). t Instantaneous velocity: Position Slope of the particle’s positiontime curve at a
given instant of time. V is tangent to x(t) when
∆t 0 Time When the velocity is constant, the average velocity over any time interval is
equal to the instantaneous velocity at any time.
Instantaneous speed: Magnitude of the instantaneous velocity.
Example: car speedometer.
 Scalar quantity
Average velocity (or average acceleration) always refers to an specific
time interval.
Instantaneous velocity (acceleration) refers to an specific instant of time. 4 III. Acceleration Average acceleration: Ratio of a change in velocity ∆v to the time interval
∆t in which the change occurs.
aavg = v2 − v1 ∆v
=
t 2 − t1 ∆t (2.5)  Vector quantity V t  Dimensions [L]/[T]2, Units: m/s2
 The average acceleration in a “vt” plot is the slope
of a straight line connecting points corresponding to
two different times.
t t Instantaneous acceleration: Limit of the average acceleration as ∆t
approaches zero.
 Vector quantity ∆v dv d 2 x
=
=
∆t → 0 ∆t
dt dt 2 a = lim (2.6)  The instantaneous acceleration is the slope of the tangent line (vt plot) at
a particular time. (green line in B)
 Average acceleration: blue line.
 When an object’s velocity and
acceleration are in the same
direction (same sign), the object is
speeding up.
 When an object’s velocity and
acceleration are in the opposite
direction, the object is slowing
down. 5  Positive acceleration does not necessarily imply speeding up, and negative
acceleration slowing down.
Example (1): v1= 25m/s ; v2= 0m/s in 5s particle slows down, aavg= 5m/s2  An object can have simultaneously v=0 and a≠0
Example (2): x(t)=At2 v(t)=2At a(t)=2A ; At t=0s, v(0)=0 but a(0)=2A Example (3):  The car is moving with constant positive velocity (red arrows maintaining
same size)
Acceleration equals zero.
Example (4):
+ acceleration
+ velocity
 Velocity and acceleration are in the same direction, “a” is uniform (blue
arrows of same length)
Velocity is increasing (red arrows are getting
longer). Example (5):
 acceleration
+ velocity
 Acceleration and velocity are in opposite directions.
 Acceleration is uniform (blue arrows same length).
 Velocity is decreasing (red arrows are getting shorter). 6 IV. Motion in one dimension with constant acceleration  Average acceleration and instantaneous acceleration are equal.
a = aavg = v − v0
t −0  Equations for motion with constant acceleration:
v = v 0 + at
v avg
v avg ( 2 .7 )
t x − x0
=
→ x = x 0 + v avg t
( 2 .8 )
t
v +v
at
=0
and ( 2 . 7 ) → v avg = v 0 +
2
2 ( 2 . 8 ), ( 2 . 9 ) → x a =x 0 = v − t +
− a = v0 v
0 avg t−0 at
2 ( 2 .9 ) 2 ( 2 .10 ) 2
2
( 2 . 7 ), ( 2 . 10 ) → v 2 = v 0 + a 2 t 2 + 2 a ( v 0 t ) = v 0 + a 2 t 2 + 2 a ( x − x 0 −
2
→ v 2 = v0 + 2 a ( x − x0 ) at 2
)
2 t ( 2 . 11 )
t missing t PROBLEMS  Chapter 2
P1. A red car and a green car move toward each other in adjacent lanes and parallel to
The xaxis. At time t=0, the red car is at x=0 and the green car at x=220 m. If the red car has
a constant velocity of 20km/h, the cars pass each other at x=44.5 m, and if it has a constant
velocity of 40 km/h, they pass each other at x=76.6m. What are (a) the initial velocity, and (b)
the acceleration of the green car? vr2=40km/h
vr1=20km/h 3 km 1h 10 m = 11.11m / s 40
⋅
⋅
h 3600s 1km Xr2=76.6m x O
Xr1=44.5 m d=220 m 44.5m
x = v t →t =
= 8s
r1 r1 1
1 5.55m / s
76.6m
x = v t → t2 =
= 6.9s
r2 r2 2
11.11m / s Xg=220m x = x +v t
r
r0 r
1
x = x + v t + at 2
g
g0 g0 2 (1)
(2) x − x g = −v t − 0.5 ⋅ a g t2 2 → 76.6 − 220 = −v ⋅ (6.9s) − 0.5 ⋅ (6.9s) 2 a g
r2
g0 2
g0
x − x g = −v t − 0.5 ⋅ ag t12 → 44.5 − 220 = −v ⋅ (8s) − 0.5 ⋅ (8s) 2 a g
r1
g0 1
g0 The car moves to the left () in my
reference system
a<0, v<0 ag = 2.1 m/s2
v0g = 13.55 m/sc 7 P2: At the instant the traffic light turns green, an automobile starts with a constant
acceleration a of 2.2 m/s2. At the same instant, a truck, traveling with constant speed of
9.5 m/s, overtakes and passes the automobile. (a) How far beyond the traffic signal will
the automobile overtake the truck? (b) How fast will the automobile be traveling at that
instant?
ac = 2.2 m/s2, vc0=0 m/s
t = 0s t = 0s x (m) Car
Truck x (m) x=0 x=d ? vt = 9.5 m/s
xT = d = vT t = 9.5 t → (1) (a) 9.5 ⋅ t = 1.1⋅ t 2 → t = 8.63 s → d = (9.5m / s )(8.63s) ≈ 82m Truck 1
xc = d = v t + act 2 → d = 0 + 0.5 ⋅ (2.2m / s 2 ) ⋅ t 2 = 1.1t 2 (2) Car
C0
2 2
(b) v 2 = v0 + 2 ⋅ ac ⋅ d = 2 ⋅ (2.2m / s 2 ) ⋅ (82m) → v f = 19m / s
f P3: A proton moves along the xaxis according to the equation: x = 50t+10t2, where x is in
meters and t is in seconds. Calculate (a) the average velocity of the proton during the first 3s of
its motion.
v avg = x (3) − x (0) (50)(3) + (10)( 3) 2 − 0
=
= 80 m / s.
∆t
3 (b) Instantaneous velocity of the proton at t = 3s. v (t ) = (c) Instantaneous acceleration of the proton at t = 3s. dx
= 50 + 20 t → v(3s ) = 50 + 20 ⋅ 3 = 110 m / s
dt a (t ) = dv
= 20 m / s 2 = a(3s )
dt (d) Graph x versus t and indicate how the answer to (a) (average velocity) can be obtained
from the plot.
(e) Indicate the answer to (b) (instantaneous velocity) on
the graph.
(f) Plot v versus t and indicate on it the answer to (c). x = 50t + 10t2
v = 50 + 20t P4. An electron moving along the xaxis has a position given by: x = 16t·exp(t) m, where t is in
seconds. How far is the electron from the origin when it momentarily stops?
x(t) when v(t)=0?? dx
= v = 16e −t − 16te −t = 16e −t (1 − t )
dt v = 0 → (1 − t ) = 0; (e−t > 0) → t = 1s x(1) = 16 / e = 5.9m 8 P5. When a high speed passenger train traveling at 161 km/h rounds a bend, the engineer is
shocked to see that a locomotive has improperly entered into the track from a siding and
is a distance D= 676 m ahead. The locomotive is moving at 29 km/h. The engineer of the
high speed train immediately applies the brakes. (a) What must be the magnitude of the
resultant deceleration if a collision is to be avoided? (b) Assume that the engineer is at x=0
when at t=0 he first spots the locomotive. Sketch x(t) curves representing the locomotive
and high speed train for the situation in which a collision is just avoided and is not quite
avoided. Locomotive Train
t = 0s x (m) dL x (m) t > 0s
x=0 vT=161km/h = 44.72 m/s = vT0 x=D x=D+dL 1D movement with a<0=cte vL=29 km/h = 8.05 m/s is constant
d L = vLt = 8.05 t → t = dL
8.05 (1) Locomotive 1
1
d L + D = v t + aT t 2 → d L + 676 = 44.72 t + aT t 2 (2) Train
T0
2
2 P5.
vTf = vT 0 + aT t = 0 → aT = −44.72m / s
(−44.72m / s)(8.05m / s) −360m2 / s 2
= (eq. 1) =
=
t
dL
dL v2Tf = v 2T 0 + 2aT ( D + d L ) = 0 → aT = −(44.72m / s)2
2(676m+d L ) (3) (4) (3) = (4) → d L = 380.3m from (1) → t =
(1) + (3) → aT = Collision can be avoided
Locomotive dL 8.05 = 47.24s −360m2 / s 2
= −0.947m / s 2
380.3m x L = 676 + 8 . 05 t
x T = 44 . 72 t + 0 . 5 a T t 2
 Collision can be avoided: Train Collision can not be avoided Slope of x(t) vs. t locomotive at t = 47.24 s (the point were both
Lines meet) = v instantaneous locom > Slope of x(t) vs. t train
 Collision cannot be avoided:
Slope of x(t) vs. t locomotive at t = 47.24 s < Slope of x(t) vs. t train 9  The motion equations can also be obtained by indefinite integration:
dv = a dt → ∫ dv = ∫ a dt → v = at + C ; v = v0 at t = 0 → v0 = (a )(0) + C → v0 = C → v = v0 + at 1
dx = v dt → ∫ dx = ∫ v dt → ∫ dx = ∫ (v0 + at )dt → ∫ dx = v0 ∫ dt + a ∫ t dt → x = v0t + at 2 + C ' ;
2
1
12
x = x0 at t = 0 → x0 = v0 (0) + a(0) + C ' → x0 = C ' → x = x0 + v0t + at
2
2 V. Free fall Motion direction along yaxis ( y >0 upwards)
Free fall acceleration: (near Earth’s surface)
a= g = 9.8 m/s2 (in mov. eqs. with constant
acceleration)
Due to gravity
Earth’s center downward on y, directed toward Approximations:
 Locally, Earth’s surface essentially flat
at slightly different points. free fall “a” has same direction  All objects at the same place have same free fall “a” (neglecting air influence). VI. Graphical integration in motion analysis From a(t) versus t graph
integration = area between
acceleration curve and time axis, from t0 to t1 v(t)
t1 v1 − v0 = ∫ a ⋅dt
t0 Similarly, from v(t) versus t graph
under curve from t0 to t1 x(t) integration = area t1 x1 − x0 = ∫ v ⋅dt
t0 10 P6: A rocket is launched vertically from the ground with an initial velocity of 80m/s. It ascends
with a constant acceleration of 4 m/s2 to an altitude of 10 km. Its motors then fail, and the
rocket continues upward as a free fall particle and then falls back down.
(a) What is the total time elapsed from takeoff until the rocket strikes the ground?
(b) What is the maximum altitude reached?
1) Ascent
a0= 4m/s2
(c) What is the velocity just before hitting ground?
y1 − y0 = v t + 0.5 ⋅ a0t12 → 104 = 80t + 2t12 → t = 53.48s y 01 y2=ymax
a1= g t2 t3=t2 2) Ascent +v1, t1 y1= 10 km
a0= 4m/s2 t1 t4 v0= 80m/s a1 = − g = a2= g 1 a= 9.8 m/s2 0−v1
− 294m / s
→ t2 =
= 29.96s
t2
−9.8m / s 2 Total time ascent = t1+t2= 53.48 s+29.96 s= 83.44 s t0=0 v3 1 vv
a0 = 1− 0 → v1 = (4m / s 2 ) ⋅ (53.48s) + 80m / s = 294m / s
t1 v2=0, t2 x 3) Descent a= 9.8 m/s2
4 0 − y1 = −v1t + 0.5 ⋅ a0t4 2 → −10 = −294t4 − 4.9t4 2 → t = 24.22 s
4 4 ttotal=t1+2t2+t4= 53.48 s + 2·29.96 s + 24.22 s=137.62 s
hmax= y2 y2104 m = v1t24.9t22= (294 m/s)(29.96s)(4.9m/s2)(29.96s)2 = 4410 m a2 = − g = hmax=14.4 km v3−(−v )
1 → v = − g ⋅ t − v = −531.35m / s
3
41
t4 11 ...
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This note was uploaded on 07/30/2011 for the course PHY 2049 taught by Professor Saha during the Spring '08 term at University of Central Florida.
 Spring '08
 SAHA
 Physics, Acceleration, Mass

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