MOTION why - Florida Atlantic University
Transcript of MOTION why - Florida Atlantic University
CHAPTER 2
MOTION IN ONE DIMENSION (KINEMATICS)
• Why objects move
• Displacement and velocity • Average velocity and average speed • Relative velocity • Instantaneous velocity • Displacement-time graphs
• Acceleration • Average acceleration • Instantaneous acceleration • Velocity-time graphs • Equations of motion
• Free fall
This chapter is about MOTION ... so, why do objects move?
It is FORCE that causes a change in motion (starting, stopping, going round corners, etc).
It is the force of the road on the tires that cause a car to move!
It is the force of gravity downward that makes the cat fall! (Its weight.)
It is the force exerted by the typists fingers that makes the keys move!
For the moment we won’t worry about what causes motion but the rules that tell us how objects move.
Displacement and distance travelled
Displacement is . Distance travelled is also.
Displacement is . Distance travelled is .
Displacement is . Distance travelled is .
Note: • distance travelled displacement, • displacement can be positive or negative, • distance always .
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�
(x2 − x1) = Δx
(x2 − x1)
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(x3 − x1)
(x3 − x1)
(x3 − x1) = (d1 + d2)
(d1 + d2)
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≥
≥ 0
Average velocity and average speed
• Average velocity .
• Average speed .
In this example, ,
i.e., average speed > average velocity
Average speed always Average velocity can be
Dimension: velocity and speed
Units: .
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Δt = t2 − t1
vav = displacement
time interval= Δx
Δt
= distance
time interval=
d1 + d2Δt
(d1 + d2) > (x3 − x1)
⇒ [L]
[T]
m/s, km/h, ft/s, mi/h, etc.
> 0. ≤ 0.
Question 2.1: A runner runs for 2.5 km on a straight path
in 9.0 mins and takes 30.0 mins to walk back to the
starting point.
(a) What is the runner’s velocity for
(b) What is the average velocity for the time spent walking?
(c) What is the average speed for the whole trip?
(d) What is the average velocity for the whole trip?
(e) Draw a position-time graph for the problem.
0 > t ≥ 9 mins? vav =
(x2 − x1)Δt1
= 2.50 km9 min
(a) From :
(b) From :
(c) Average speed
(d)
x1 → x2
= 2500 m
9 × 60 s= 4.63 m/s.
x2 → x1 vav =
(x1 − x2)Δt2
= −2.50 km30 min
= −2500 m
30 × 60 s= −1.39 m/s.
= total distance
total time= 5000 m
39 × 60 s= 2.14 m/s.
vav = displacement
total time= 0.
(e) The position-time graph for the problem. The runner’s
displacement is the position relative to the starting point.
• The slope of AB is , i.e., the
average velocity for the first 9 mins.
• The slope of BC is , i.e., the
average velocity while walking.
Important concept: the average velocity is the slope of the
displacement-time graph.
2.50 km9 min
⇒ 4.63 m/s
−2.50 km
30 min⇒−1.39 m/s
Relative velocity
If the velocity of the walkway relative to the ground (i.e.,
the stationary observer) is vwg, and the velocity of 2
with respect to 1 is v21, then the velocity of 2 with
respect to a stationary observer (the relative velocity) is:
(vwg + v21)
Note, if 2 is walking in the opposite direction, his
velocity relative to the observer is
(vwg − v21)
In that case, what “happens” if vwg = v21 ?
This is called a Galilean transformation.
vwg
v21 1 2
Question 2.2: Two trains, 35 km apart and traveling
at 10 km/h, are approaching each other on parallel
tracks. A bird flies back and forth between the trains
at 15 km/h relative to the ground until the trains pass
each other.
(a) How long is the bird flying?
(b) What distance does the bird fly?
35 km
10 km/h 10 km/h
Assume the bird flies a total distance d.
Av. speed of bird is
where t is the elapse time before the trains meet.
Imagine you are sitting on the front of the left hand
train, then the relative speed of approach of the right
hand train to you is
So the time taken for the trains to pass is
distance flown
time= d
t= 15 km/h,
i.e., d = (15 km/h) × t,
(10 km/h) + (10 km/h) = 20 km/h.
t = distance they travel
relative speed= 35 km
20 km/h= 1.75 h.
∴d = (15 km/h)(1.75 h) = 26.3 km.
35 km
10 km/h 10 km/h
Question 2.3: A river flows from left to right with a
velocity of 0.20 m/s relative to the bank. If you swim
with a velocity of 1.00 m/s relative to the water, does it
take you more time, less time, or the same time to swim
from than if the river was not flowing at all? A → B→ A
When you swim from your velocity relative to
the bank is
When you swim from your velocity relative to
the bank is
If the distance AB is d meters, then the times taken (in
seconds) are:
So the total elapsed time is
If the river is not flowing then
i.e., it takes more time if the river is flowing!
A → B
vAB = 1.00 m/s + 0.20 m/s = 1.20 m/s.
B→ A
vBA = 1.00 m/s − 0.20 m/s = 0.80 m/s.
tAB = d
1.20 m/s and tBA = d
0.80 m/s
tAB + tBA = (1.20 m/s + 0.80 m/s)d
(0.80 m/s)(1.20 m/s)= 2.08d s.
′t = d
1.00 m/s+ d
1.00 m/s= 2.00d s,
Instantaneous velocity
Velocities are rarely constant. Consider the displacement
time graph below in which the velocity changes with time.
The average velocity
from is
Note that
We define the instantaneous velocity at the point as the
slope of the tangent at the point
i.e., the first derivative of x with respect to t at the point .
P ′2 P1 → P2
v1→2 = Δx
Δt=
x2 − x1t2 − t1
.
P1
P1
∴v(t) = LimitΔt→0ΔxΔt
= dxdt
⎡
⎣⎢
⎤
⎦⎥
P1
,
P1,
v1→2 < v1→ ′2 .Question 2.4: The plot shows the variation of the
position of a car with time.
(a) What was its velocity at ?
(b) During the time interval was the car
speeding up, slowing down or moving at constant
speed?
t = 15 s
0 < t <10 s
(a) Velocity is defined as the slope at
i.e.,
t = 15 s,
v = dx
dt⎛⎝⎜
⎞⎠⎟ t=15 s
= Δx
Δt= (2 m − 4 m)
(20 s −10 s)= −0.20 m/s.
(b) The instantaneous speed is given by the magnitude
of the slope of the displacement-time curve.
The magnitude of the slope increases over the
time interval . Therefore, the car was
speeding up.
Note, in the interval the speed was
constant. During the interval the car was
slowing down.
v = dx
dt 0 < t ≤10 s
10 s ≤ t ≤ 20 s
20 s ≤ t ≤ 30 s
Whenever velocity changes we have … acceleration.
Average acceleration
Dimension:
Units:
Instantaneous acceleration
Since then also.
aav = Δv
Δt=
v2 − v1t2 − t1
.
[L][T]
1[T]
= [L]
[T]2⇒ [L][T]−2.
m/s2, ft/s2, etc.
a = LimitΔt→0
ΔvΔt
= dvdt
.
v = dx
dt,
a = d2x
dt2
Equations of motion
To start with, assume that the acceleration a is constant.
Since we can integrate this expression with
respect to t and obtain:
where is an integration constant. If the initial
velocity (when ) is , then i.e.,
Also, since
where is another constant. If when
then so
i.e., the displacement is
dvdt
= a,
v = dv = a∫∫ dt = at +C1,
C1
t = 0 v! C1 = v!,
v = v! + at ... ... ... ... (1)
v = dx
dt,
x = dx∫ = vdt = (v! + at)dt∫∫
= v!t +
12
at2 +C2,
C2 x = x! t = 0,
C2 = x!,
x = x! + v!t +
12
at2,
(x − x!) = v!t +
12
at2 ... ... ... (2)
Since the acceleration is constant we can define the
average velocity
Then the displacement is
But, from (1)
and substituting for t in (4) we find
Equations (1) – (5) are the principle kinematic equations
for one-dimensional motion at constant acceleration.
vav = 1
2(v + v!) ... ... ... ... (3)
Δx = x − x! = vavt = 1
2(v + v!)t ... (4)
t =
(v − v!)a
,
x − x! =
12
(v + v!)(v − v!)
a
=
(v2 − v!2)
2a.
∴v2 = v!2 + 2a(x − x!) ... ... ... (5)
If the acceleration is not constant, providing we know how
the velocity varies with time, we can still determine the
displacement over a time interval. For example, given
the velocity-time variation shown above, what is the
displacement of an object over the time period
We split the time period into a
large number of equal time
intervals, If is very small,
the displacement from to
is so the total displacement from is
t1 → t2 ?
Δt. Δt
t t + Δt
Δx ≈ viΔt, t1 → t2
x(t2 ) − x(t1) ≈ v1Δt + v2Δt +! ≈ (viΔti∑ ).
The sum becomes exact if Then
i.e., the displacement is the area under the velocity-time
plot from to So, if we know how the velocity
varies with time, we can determine the
displacement over any time interval.
Note, the above expression is valid even if the
acceleration is changing, when our previous equations
of motion (1) – (5) cannot be applied.
Δt → 0.
x(t2 ) − x(t1) = LimitΔt→0 viΔti∑( ) = vdt,
t1
t2
∫
t1 t2.
i.e., v → v(t),
Question 2.5: A car starts from rest at and
accelerates at a constant rate of for 5.0 s. It
then travels at a constant speed for 15.0 s, when the
driver is forced to slam on the brakes. The car comes to
stop after a distance of 8.0 m.
(a) What is the velocity of the car after the initial 5.0 s?
(b) After the driver slams on the brakes and the car
comes to a stop, what is the acceleration?
(c) When the car comes to stop, how far is it from the
starting point?
x! = 50.0 m
4.0 m/s2
t = 0 t1 t2 t3
The accelerations from are
constant, so we can use eqs. (1) – (5).
(a) Using (1) we have
(b) To find the acceleration we use eq. (5):
where d is distance traveled from .
The acceleration is negative because the car is slowing.
(c) Here are two ways to find the total distance.
• Find the distance traveled over each time interval.
• Plot the velocity-time graph and find the area.
v = v! + at = 0 + (4.0 m/s2)(5.0 s) = 20.0 m/s.
t2 → t3
v2 = v!
2 + 2ad ⇒ a =v2 − v!
2
2d
∴ a = 0 − (20.0 m/s)2
2(8.0 m)= −25.0 m/s2.
t = 0 to t1 and from t2 to t3
From and using eq. (2) we have
Since the velocity is constant from , the distance
traveled is
From the distance traveled is 8.0 m, so the total
distance traveled by the car is
To confirm this result, we plot the velocity-time graph,
but we need , i.e., the time for the car to stop.
Using eq. (1) we find
Here is the velocity-time graph …
t → t1
x − x! = v!t +
12
at2 = 0 + 12
(4.0 m/s2)(5.0 s)2 = 50.0 m.
t1 → t2
v(t2 − t1) = (20.0 m/s)(15.0 s) = 300 m.
t2 → t3
(50.0 m) + (300 m) + (8.0 m) = 358 m.
(t3 − t2 )
(t3 − t2 ) =
v − v!a
= 0 − 20.0 m/s
−25.0 m/s2= 0.8 s.
time (s)
velocity
20.0 (m/s)
0 5 20 20.8
A1 A2 A3
The total distance traveled is given by the area under
the velocity-time graph, i.e., A1 + A2 + A3
= 12
(5.0 s)(20.0 m/s) + (15.0 s)(20.0 m/s)
+ 12
(0.8 s)(20.0 m/s)
= 358 m.
Question 2.6: An object, initially at rest at
experiences an acceleration given by
After any time t, what is
(a) the instantaneous velocity, and
(b) the position of the object relative to the origin?
(c) What is the displacement of the object over the
time period
x! = 1.00 m,
a(t) = 1.50 + 0.20t m/s2.
2.0 s ≥ t ≥ 4.0 s?
Because the acceleration is not constant we cannot use eqs.
(1) – (5); we must use calculus.
(a) Given then
Since the initial velocity is zero at then .
(b) Since the position is given by
The initial position at is then
a(t) = dv
dt= 1.50 + 0.20t,
v = dv = (1.50 + 0.20t)dt∫∫
= 1.50t + 0.20 12
t2⎛⎝⎜
⎞⎠⎟+C1.
t = 0, C1 = 0.
∴v(t) = 1.50t + 0.10t2 m/s.
v = dx
dt,
x = dx∫ = vdt =∫ 1.50t + 0.10t2( )dt∫
= 1.5012
t2⎛⎝⎜
⎞⎠⎟+ 0.10
13
t3⎛⎝⎜
⎞⎠⎟+C2.
t = 0 x! = 1.00 m, C2 = 1.00 m.
∴x(t) = 1.00 + 0.75t2 + 0.033t3 m.
(c) By definition, the displacement is
Using the result in part (b)
Check using (b):
Δxt1→t2
= vdt =t1
t2∫ 1.50t + 0.10t2( )
2
4∫ dt.
1.50t + 0.10t2( )2
4∫ dt = 0.75t2 + 0.033t3 +C2
⎡⎣⎢
⎤⎦⎥2
4
= 0.75 42 − 22( ) + 0.033 43 − 23( )= 10.85 m.
x(t = 2) = 1.00 + 0.75(22) + 0.033(23) = 4.26 m.
x(t = 4) = 1.00 + 0.75(42) + 0.033(43) = 15.11 m.
∴Δx = (15.11− 4.26)m = 10.85 m.
Question 2.7: An object, initially at rest at the origin,
experiences an acceleration
(a) What is its speed when
(b) How far from the origin is it at that time?
a(t) = 1.50e−0.3t m/s2.
t = 5 s?
(a)
But
(b)
We have a(t) = dv(t)dt
= 1.50e−0.3t m/s2.
∴v(t) = dv(t)∫ = 1.50e−0.3t∫ dt
= − 1.50.3
e−0.3t +C1 = −5e−0.3t +C1.
We have v(t) = dx(t)dt
= 5 1− e−0.3t( ) m/s.
∴x(t) = dx(t) = 5 1− e−0.3t( )∫∫ dt = 5 t + e−0.3t
0.3
⎛
⎝⎜
⎞
⎠⎟ +C2.
But x(0) = 0 so C2 = −16.67.
∴x(t) = 5 t + e−0.3t
0.3
⎛
⎝⎜
⎞
⎠⎟ −16.67 m.
When t = 5 s, x(5s) = 5 5+ e−1.5
0.3
⎛
⎝⎜
⎞
⎠⎟ −16.67
= 12.05 m.
v(0) = 0 so C1 = 5. ∴v(t) = 5 1− e−0.3t( ) m/s.
When t = 5 s, v(5 s) = 5 1− e−1.5( )= 5(1− 0.223) = 3.88 m/s.
Free fall – Galileo Galilei (1564-1642)
In “free fall”, i.e., in the absence of air
resistance, all objects close to the
Earth’s surface fall with a constant
acceleration (directed downward)
known as the “acceleration due to
gravity”.
Therefore, the velocity of the object in free fall increases
by 9.81 m/s each second. For an object dropped from
rest at a height its velocity and vertical displacement
are
and
Note that neither the velocity nor the displacement
depend on the mass of the object!
a ⇒ g = 9.81 m/s2 (32.2 ft/s2 )
y! : t = 0
y : t
y!,
y − y! = v!t +
12
at2 = 12
gt2.
v = v! + at = gt
So, in the absence of air resistance, all objects fall at the
same rate:
and all objects fall the same distance in the same time:
The velocity of an object in free-fall, starting from rest.
velo
city
(m/s
) D
ista
nce
(m) The distance fallen by
an object in free-fall, starting from rest.
Time elapse photographs of an
apple and feather released at
the same time, in the absence of
air resistance (free fall).
Apollo 15 astronaut David Scott drops a geological
hammer (1.32 kg) and a falcon feather (0.03 kg) at the
same time on the Moon (July/August 1971).
Question 2.8: A ball is dropped from rest from the top of
a tower 20.0 m high. A second ball is thrown vertically
upwards from the ground at the same instant the first ball
is released. If the balls pass each other at the half-height
point, i.e., a height of 10.0 m above the ground,
(a) how long does each ball take to reach the half-
height point?
(b) What is the initial speed of the second ball?
(a) Since the balls are released at the same time, when
they pass, they will have traveled for the same time
interval. We take the upward direction as the positive
direction. So, for the first ball the mid way point is at
To find the time to reach the midway point we
use
i.e.,
(b) The time for the second ball to reach the half-height
point is the same but the displacement of the second ball
is upwards. For the second ball,
(y − y!) = v!t +
12
at2, t =
2(y − y!)a
y = 10 m.
= 2(10 m − 20 m)
−9.81 m/s2= 1.43 s.
(y − y!) = v!t +
12
at2 = v!t −12
gt2.
∴v! =
(y − y!)t
+ 12
gt = 10 m1.43 s
+ 12
(9.81 m/s2)(1.43 s)
= 14.0 m/s.
Question 2.9: A rocket is fired vertically with a constant
upward acceleration of After 25.0 s, the engine
is shut off but the rocket continues to rise until it reaches
its maximum height. It then falls back to the ground.
Neglecting air resistance, find
(a) the height the rocket reaches,
(b) the time to reach maximum height,
(c) the velocity of the rocket just before it strikes the
ground.
20.0 m/s2.
Take the upward direction as positive.
(a)
i.e.,
Also
Between and (free fall).
i.e.,
Maximum height:
(y1 − y!) = v!(t1 − t!) +
12
a(t1 − t!)2,
y1 =
12
(20.0 m/s2)(25.0 s)2 = 6250 m.
v1 = v! + at = (20.0 m/s2)(25.0 s) = 500 m/s.
y1 y2, a = −g = −9.81 m/s2
v22 = v1
2 + 2a(y2 − y1),
(y2 − y1) =
v22 − v1
2
2a= 0 − (500 m/s)2
2(−9.81 m/s2)= 12700 m.
y2 = 6250 m +12700 m = 18950 m.
y! = 0 : v! = 0 : t! = 0
y1 : v1 : t1 = 25.0 s
y2 : v2 = 0 : t2
a = 20.0 m/s2
Free fall: a = −g
y3 = 0 : v3
(b) Total climb time is But
i.e.,
(c) i.e.,
The appropriate solution is , as we took
the upward direction as positive but the velocity is
directed downward.
t1 + t2.
v2 = v1 + a(t2 − t1),
(t2 − t1) =
v2 − v1a
= 0 − (500 m/s)
−9.81 m/s2= 51.0 s.
∴ t2 = 25.0 s + 51.0 s = 76.0 s.
v32 = v2
2 + 2a(y3 − y2) v32 = 2(−g)(−y2) = 2gy2
= 2(9.81 m/s2)(18950 m) = 3.72 ×105 (m/s)2.
∴v3 = ±610.2 m/s.
v3 = −610.2 m/s
y! = 0 : v! = 0 : t! = 0
y1 = 6250 m : v1 = 500 m/s : t1 = 25.0 s
y2 = 18950 m : v2 = 0 : t2
a = 20.0 m/s2
Free fall: a = −g
y3 = 0 : v3
Question 2.8: If, in the photograph, the upper ball is at the top of its trajectory, estimate the time it takes for a ball to loop from one hand to the other.
DISCUSSION PROBLEM [2.1]:
In the photograph, the upper ball is at the top of its
trajectory. Estimate the time it takes for a ball to loop
from one hand to the other. It is about
A 0.5 s B 0.75 s C 1.0 s D 1.25 s?
Galileo investigated the acceleration of gravity by rolling
balls down a smooth incline. So, how is the acceleration
due to gravity affected by the slope of a frictionless
incline?
On a vertical surface the acceleration is the same as in free fall, i.e.,
(θ = 90!),
a = g.
On a horizontal surface gravity does not produce an acceleration along the surface, i.e.,
(θ = 0),
a = 0.
On a surface inclined at an angle to the horizontal, the acceleration is
θ
a = gsinθ.