newton’s lawsconcepts of motion phy1012f kinematics gregor leigh [email protected]
TRANSCRIPT
NEWTON’S LAWS CONCEPTS OF MOTIONPHY1012F
2
KINEMATICS
Learning outcomes:At the end of this chapter you should be able to…
Interpret, draw and convert between position, velocity and acceleration graphs for 1-d motion.
Use an explicit problem-solving strategy for kinematics problems.
Apply appropriate mathematical representations (equations) in order to solve numerical kinematics problems involving motion in one dimension.
NEWTON’S LAWS CONCEPTS OF MOTIONPHY1012F
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The positive end of the x-axis points to the right; The positive end of the y-axis points upwards.
MOTION IN ONE DIMENSION
We shall standardise on the following sign conventions:
y
x
NEWTON’S LAWS CONCEPTS OF MOTIONPHY1012F
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x < 0; y > 0
Positions left of the y-axis have negative x values; Positions right of the y-axis have positive x values.Positions below the x-axis have negative y values;Positions above the x-axis have positive y values.
MOTION IN ONE DIMENSION
We shall standardise on the following sign conventions:
y
x
x > 0; y > 0
0x = 0; y > 0
x > 0; y < 0x < 0; y < 0
x < 0; y = 0
NEWTON’S LAWS CONCEPTS OF MOTIONPHY1012F
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Vectors pointing to the right (or up) are positive; Vectors pointing to the left (or down) are negative.
MOTION IN ONE DIMENSION
We shall standardise on the following sign conventions for representing directions:
y
x
0v
0a
NB: The signs represent the directions. The magnitudes of vectors can never be negative!
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MOTION IN ONE DIMENSION
In 1-d the relationship between acceleration and velocity simplifies to…
When is zero, velocity remains constant.
If and have the same sign, the object is speeding up.
If and have opposite signs, the object is slowing down.
a
a
v
a
v
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The slope at a point on a position-vs-time graph of an object is
A the object’s speed at that pointB the object’s acceleration at that point C the object’s average velocity at that point D the object’s instantaneous velocity at that pointE the distance travelled by the object to that point
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POSITION GRAPHS
Plotting a body’s position on a vertical axis against time on the horizontal axis produces a position-vs-time graph, or position graph…
v 1 frame per minute
x (m)0 200100 300 400
x (m)
t (min)
200
400
2 4 60
0
NEWTON’S LAWS CONCEPTS OF MOTIONPHY1012F
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POSITION GRAPHS
Plotting a body’s position on a vertical axis against time on the horizontal axis produces a position-vs-time graph, or position graph…
v
x (m)
t (min)
200
400
2 4 60
0
1 frame per minute
x (m)0 200100 300 400
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INTERPRETING POSITION GRAPHS
It is essential to remember that motion graphs are abstract representations of motion – they are NOT pictures!The following graph represents the motion of a car along a straight road… Describe the motion of the car.
x (km)
t (min)
–10
20 40 600
–20
10
80
After 30 min the car stops for 10 min at a position 20 km to the left of the origin.The car reaches the origin once more at 80 min.At 40 min the car starts moving back to the right.During the first 30 min the value of x changes from +10 km to 20 km, indicating that the car is moving to the left.
At t = 0 the car is 10 km to the right of the origin.
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UNIFORM MOTION
Straight-line motion in which equal displacements occur during any successive equal-time intervals is called uniform motion.
Motion diagram:
Position graph:
s (m)
t (s)60 2 4
2
4
6
0
s = 4 m
t = 6 s
avg slope of position graphv
avgsvt
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UNIFORM MOTION
Straight-line motion in which equal displacements occur during any successive equal-time intervals is called uniform motion.
Motion diagram:
Position graph:
An object’s straight line motion is uniform if and only if its velocity vx (or vy) is constant and unchanging.
avg
avg slope of position graph
svt
v
s (m)
t (s)60 2 4
2
4
6
0
t = 4 s
s = 6 m
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POSITION GRAPHS OF UNIFORM MOTION
Body B is… travelling to the left/right and is going slower/faster than A.
Body A is travelling to the right at constant speed…
x (m)
t (s)
A
x (m)
t (s)
B
(Assume same axes scales.)
NEWTON’S LAWS CONCEPTS OF MOTIONPHY1012F
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POSITION GRAPHS OF UNIFORM MOTION
Summary:
Zero slope zero velocity. (Object is stationary.)
Steeper slopes faster speeds.
Negative slope negative velocity (ie vx is left/vy is
down).
The sign (negative or positive) refers only to the direction of the velocity, and has nothing to do with its magnitude.
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POSITION GRAPHS OF UNIFORM MOTION
Summary:
Zero slope zero velocity. (Object is stationary.)
Steeper slopes faster speeds.
Negative slope negative velocity (ie vx is left/vy is
down).
The sign (negative or positive) refers only to the direction of the velocity, and has nothing to do with its magnitude.
The slope is a ratio of intervals, x/t, not coordinates, x/t.
Remember: Position graphs are abstract representations! We are concerned with the physically meaningful slope [in m/s], not the actual slope of the graph on paper.
NEWTON’S LAWS CONCEPTS OF MOTIONPHY1012F
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0 1 2 3 4
Describe carefully (and quantitatively) the motion of the basketball player depicted by this position graph.
x (m)
t (s)
2
4
6
0
NEWTON’S LAWS CONCEPTS OF MOTIONPHY1012F
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THE MATHEMATICS OF UNIFORM MOTION
The velocity of a uniformly moving object tells us the amount by which its position changes during each second.
f i ss s v t
ssvt
s (m)
t (s)
sf
si
tfti
t
s
(For uniform motion)
f i
f i
s st t
NEWTON’S LAWS CONCEPTS OF MOTION
BobBob
PHY1012F
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Bob leaves home in Chicago at 09:00 and travels east at a steady 100 km/h. Susan, 680 km to the east in Pittsburgh, leaves at the same time and travels west at a steady 70 km/h. Where will they meet?
Pictorial representation:
(x0)B, (vx)B, t0
(ax)B = 0
(x0)S, (vx)S, t0(x1)B, (vx)B, t1
(x1)S, (vx)S, t1
(ax)S = 0
MULTI-REPRESENTATIONAL PROBLEM-SOLVING
(x0)B = 0 km(x0)S = +680 km
(vx)B = +100 km/h(vx)S = –70 km/h
t0 = 0 h t1 is when (x1)B = (x1)S
(x1)B = ?
x0
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MULTI-REPRESENTATIONAL PROBLEM-SOLVING
Bob leaves home in Chicago at 09:00 and travels east at a steady 100 km/h. Susan, 680 km to the east in Pittsburgh, leaves at the same time and travels west at a steady 70 km/h. Where will they meet?
Physical representation:
Bv
Chicago
Sv
PittsburghmeetB 0a
S 0a
NEWTON’S LAWS CONCEPTS OF MOTIONPHY1012F
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Bob leaves home in Chicago at 09:00 and travels east at a steady 100 km/h. Susan, 680 km to the east in Pittsburgh, leaves at the same time and travels west at a steady 70 km/h. Where will they meet?
Graphical representation:
MULTI-REPRESENTATIONAL PROBLEM-SOLVING
x (km)
200
0tmeet
t (h)0
400
600Susan
Bob
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680 km
100 km/h 70 km/h
Bob leaves home in Chicago at 09:00 and travels east at a steady 100 km/h. Susan, 680 km to the east in Pittsburgh, leaves at the same time and travels west at a steady 70 km/h. Where will they meet?
Mathematical representation:
(x1)B = (x0)B + (vx)B(t1 – t0)
(x1)S = (x0)S + (vx)S(t1 – t0)
They meet when (x1)B = (x1)S
(x1)B = (vx)Bt1 = 100 km/h 4.0 h = 400 km east of Chicago
MULTI-REPRESENTATIONAL PROBLEM-SOLVING
sf = si + vst
= (vx)Bt1
= (x0)S + (vx)St1
i.e. (vx)Bt1 = (x0)S + (vx)St1
0 S1
B Sx x
xt
v v
4.0 hours
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INSTANTANEOUS VELOCITY
Adjusting the time interval between “movie frames” of the horizontally orbiting tennis ball alters the average velocity vectors and the information they convey…
1v
0v
2v3v
4v
5v
6v
7v
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INSTANTANEOUS VELOCITY
Adjusting the time interval between “movie frames” of the horizontally orbiting tennis ball alters the average velocity vectors and the information they convey…
0v
1v
2v
3vThe longer the time
interval, the less “real” the representation:(Especially if the “background” information is removed.)
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INSTANTANEOUS VELOCITY
Conversely, the shorter the time interval, the more the vectors tend to show us what the ball is really doing – rather than merely depicting “average” behaviour for that time interval.
As t gets smaller and smaller, approaches a limit – a constant value representing the instantaneous velocity at that point in time.
avgs
tv
0limst
s dsvt dt
Mathematically:
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ctn1. Multiply the expression by the existing
index.
2. Subtract 1 from the index.
n –1dudt
DERIVATIVES
0limst
s dsvt dt
In general (using an arbitrary function as a template), if u = ctn, to find the derivative of u (with respect to t)…
d du dwu wdt dt dt
The derivative of a sum is the sum of the derivatives:
dsdtThe limit, , is called
the derivative of s with respect to t.
E.g.
212s ut at
i.e.
v u at
1 1 2 1121 2ds ut at
dt
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INSTANTANEOUS VELOCITY
The same process of shrinking the time interval to determine the instantaneous velocity at one particular time can also be applied to linear motion.
In this case, however, it is more helpful to make use of a position graph rather than a motion diagram…
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12 14 164 6 8
INSTANTANEOUS VELOCITY
The motion diagram represents an object whose speed is NOT constant, but increases uniformly each second:v 1 frame per second
s (m)0 2 10
(Plotting uniformly accelerated motion against time results in a parabolically-shaped position graph.)
2 4 60t (s)
8
4
8
0
s (m)
12
16
s
s
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INSTANTANEOUS VELOCITY
s (m)
t (s)
The ratio gives the average velocity, , for that particular time interval, represented graphically by the slope of the dotted line.
st
avgv
The larger t, the less detailed the information…
s
s s
ttt
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INSTANTANEOUS VELOCITY
s (m)
t (s)
Conversely, if we let the time interval either side of time t shrink towards the limit (t0), we get the instantaneous velocity at time t.
On a position graph this corresponds to the slope of the tangent to the curve at time t.
t
s
t0
limst
s dsvt dt
Mathematically:
NEWTON’S LAWS CONCEPTS OF MOTIONPHY1012F
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INSTANTANEOUS vs AVERAGE VELOCITY
s (m)
t (s)
Note that (for uniform acceleration) the average velocity for a whole time interval is the same as
t
the instantaneous velocity at time t in the middle of the interval…
…as illustrated by the parallel slopes of the dotted lines on the position graph.
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xt
vx (m/s)
–10
2 4 60
5
8t (s)
For the first 3 s the
is
POSITION GRAPHS VELOCITY GRAPHS
Velocity is equivalent to the slope of a position graph.
x (m)
t (s)
–102 4 6
0
–20
10
8
20 10 m 10 m/s3 0 s
xt
E.g. A car travels along a straight road…
velocityslope
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Between 3 s and 4 s the
is
POSITION GRAPHS VELOCITY GRAPHS
Velocity is equivalent to the slope of a position graph.
x (m)
–102 4 6
0
–20
10
8
vx (m/s)
–10
2 4 60
5
8
20 20 m0 m/s
1 sxt
E.g. A car travels along a straight road…
t (s)
t (s)
velocityslope
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x
t
Between 4 s and 8 s the
is
POSITION GRAPHS VELOCITY GRAPHS
Velocity is equivalent to the slope of a position graph.
x (m)
–102 4 6
0
–20
10
8
vx (m/s)
–10
2 4 60
5
8
0 20 m5 m/s
8 4 sxt
E.g. A car travels along a straight road…
t (s)
t (s)
velocityslope
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For the first 3 s the
increases steadily from zero to 7 m/s.
POSITION GRAPHS VELOCITY GRAPHS
Velocity is equivalent to the slope of a position graph.
x (m)
t (s)
–102 4 6
0
–20
10
8
E.g. A car travels along a straight road…
velocityslope
vx (m/s)
2 4 60
8t (s)
4
8
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From 3 s to 6 s the
remains a steady 7 m/s.
POSITION GRAPHS VELOCITY GRAPHS
Velocity is equivalent to the slope of a position graph.
x (m)
t (s)
–102 4 6
0
–20
10
8
E.g. A car travels along a straight road…
velocityslope
vx (m/s)
2 4 60
8t (s)
4
8
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Between 6 s and 7 s the
quickly decreases to 0…
POSITION GRAPHS VELOCITY GRAPHS
Velocity is equivalent to the slope of a position graph.
x (m)
t (s)
–102 4 6
0
–20
10
8
E.g. A car travels along a straight road…
velocityslope
vx (m/s)
2 4 60
8t (s)
4
8
…and remains there.
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THREE-MINUTE PAPER
On a smallish piece of paper (which you’re going to fold in half), answer the following questions:
What have you learnt so far?
What still confuses you the most?
Are you having fun? (If not, why not?!)
NEWTON’S LAWS CONCEPTS OF MOTIONPHY1012F
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FINDING POSITION FROM VELOCITY
In the previous chapter we showed that a body’s position can be determined from its velocity using .
f is s v t
Graphically, the change in position (s = vt) is given by
the area “under” a velocity graph:
tv
vs (m/s)
2 4 60
8t (s)
4
8
During the time interval 2 s to 8 s the body travels a distance
8 m/s 8 2 s
48 m
v t
s
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FINDING POSITION FROM VELOCITY
In the previous chapter we showed that a body’s position can be determined from its velocity using .
f is s v t
Graphically, the change in position (s = vt) is given by
the area “under” a velocity graph:vs (m/s)
2 4 60
8t (s)
4
8
Even if the velocity varied (uniformly) during the time interval, s could still be determined by summing the “bits”:1 2 3s s s s
s1 s3s2
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FINDING POSITION FROM VELOCITY
If the velocity varies non -uniformly during the interval…
vs (m/s)
ti tf
t (s)
The total area under the graph is approximately
…we can approximate the motion with a series of constant velocity intervals.
s1 s2
t t
(vs)1
(vs)2 1 2s ss v t v t
2
1s k
ks v t
1 2s s s
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s3
s4
s1 s2
FINDING POSITION FROM VELOCITY
Once again we apply calculus, shrinking the t’s to obtain more and more accurate approximations…
vs (m/s)
until, as t 0,
1 2 1 2N s s s Ns s s s v t v t v t
f
0 1 i
limtN
s skt k tv t v dt
t (s)ti tf
f
f ii
t
st
s s v dt So
1
N
s kk
s v t
i.e.
NEWTON’S LAWS CONCEPTS OF MOTIONPHY1012F
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INTEGRALS
is called the integral of vs dt from ti to tf.
Since it has two definite boundaries (ti and tf), it
is known as a definite integral.
f
i
t
st
v dt
In general (using an arbitrary function as a template),
The integral of a sum is the sum of the integrals:
f f f
i i i
t t t
t t tu w dt udt w dt
1 1f i
1 1
n nct ctn n
ff 1
i i1
tt nn
t t
ctct dtn
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f
i
t
t
udt +1
To integrate u = ctn …
ctn
DIFFERENTIATION AND INTEGRATION FOR DUMMIES
To differentiate u = ctn …
1. Multiply the expression by the existing index.
2. Subtract 1 from the index.
n –1dudt
ctn1. Add 1 to the index.
2. Divide the expression by the new index.
3. Evaluate the integral at the upper limit,
and...
4. subtract the lower limit value of the
integral.
n +1
f
i
t
t
1f
1
nctn
1i
1
nctn
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FINDING POSITION FROM VELOCITY
A body which starts at position xi = 30 m at time ti,
moves according to vx = (– 5t + 10) m/s.
Where does the body turn around?At what time does the body reach the origin?
2 4 6t (s)
vx (m/s)
0
10
–20
–10
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0
10t
dt0
5t
t dt
0
30 5 10t
t dt
FINDING POSITION FROM VELOCITY
A body which starts at position xi = 30 m at time ti,
moves according to vx = (– 5t + 10) m/s.
Where does the body turn around?At what time does the body reach the origin?
f i0
t
xx x v dt
and
2f
530 10 m2
x t t Substitute t = 2 and solve for x. Substitute x = 0 and solve for t.
0 0
30 5 10t t
t dt dt
2
0
52
tt 2 25 5 0
2 2t 25
2t
010
tt 10 0t 10t
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For the first 6 s the
is
VELOCITY GRAPHS ACCELERATION GRAPHS
Acceleration is equivalent to the slope of a velocity graph.
26 0 m/s 1 m/s6 0 s
vt
E.g. A car travels along a straight road…
accelerationslope3 6 9
012
t (s)
vx (m/s)
–6
6
1
ax (m/s2)
–1
–2
3 6 90
12t (s)
NEWTON’S LAWS CONCEPTS OF MOTION
accelerationacceleration
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For the last 6 s the
is
VELOCITY GRAPHS ACCELERATION GRAPHS
Acceleration is equivalent to the slope of a velocity graph.
26 6 m/s 2 m/s12 6 s
vt
E.g. A car travels along a straight road…
slope3 6 9
012
t (s)
vx (m/s)
–6
6
1
ax (m/s2)
–1
–2
3 6 90
12t (s)
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SUMMARY OF GRAPHS OF MOTION
t
s
Constant +ve velocity
t
as
0
t
vs
vis
t
s
Increasing +ve velocity
t
as
0
t
s
t
as
0
t
vs
vfs
vis
vis
t
vs
vfs
Decreasing +ve velocity
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SUMMARY OF GRAPHS OF MOTION
t
s
Constant –ve velocity
t
as
0
t
s
Increasing –ve velocity
t
as
0
t
s
t
as
0
tvs
vis
vfs
Decreasing –ve velocity
t
vs
0t
vs
vfs
vis
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KINEMATIC EQUATIONS OF CONSTANT a
By definition, vs (m/s)
t (s)
vfs
vis
tfti t
vs =
ast
f is s ss
v v va
t t
Hence: vfs = vis + ast
sf = si + area under v-graph between ti and tf .
½as(t)2
vist
Hence: sf = si + vist + ½as(t)2
vis
And, substituting t = (vfs – vis )/as: vfs2 = vis
2 + 2ass
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PROBLEM-SOLVING STRATEGY FOR CONSTANT ACCELERATION KINEMATICS PROBLEMS
Use the particle model. Make simplifying assumptions.
Draw a physical representation (motion diagram).
Draw a pictorial representation.
Draw a graphical representation if appropriate.
Use a mathematical representation (using the equations of motion with appropriately modified subscripts) to solve.
Assess your solution: Is it complete? Is it reasonable?
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A student is running at a constant speed of 5 m/s in an attempt to catch a Jammie Shuttle. When she is 11 m from the bus, it pulls away with a constant acceleration of 1 m/s2. From this point, how long does it take her to catch up to the bus if she keeps running with the same speed? Physical representation:
Wv
Jv
W 0a
Ja
x
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A student is running at a constant speed of 5 m/s in an attempt to catch a Jammie Shuttle. When she is 11 m from the bus, it pulls away with a constant acceleration of 1 m/s2. From this point, how long does it take her to catch up to the bus if she keeps running with the same speed? Physical representation:
Wv
Jv
W 0a
Ja
x
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A student is running at a constant speed of 5 m/s in an attempt to catch a Jammie Shuttle. When she is 11 m from the bus, it pulls away with a constant acceleration of 1 m/s2. From this point, how long does it take her to catch up to the bus if she keeps running with the same speed? Pictorial representation:
x (m)
0
(x0)W, (v0x)W, t0
(ax)W (x0)J, (v0x)J, t0 (x1)J, (v1x)J, t1
(x1)W, (v1x)W, t1
(ax)J
(x0)W = 0 m
(x0)J = +11 mt0 = 0 s
(v0x)W = (v1x)W = +5 m/s
(v0x)J = 0 m/s(ax)W = 0 m/s2
(ax)J = +1 m/s2
(x1)W = (x1)J = ?(v1x)J = ?
t1 = ? is when (x1)W = (x1)J
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A student is running at a constant speed of 5 m/s in an attempt to catch a Jammie Shuttle. When she is 11 m from the bus, it pulls away with a constant acceleration of 1 m/s2. From this point, how long does it take her to catch up to the bus if she keeps running with the same speed? Graphical representation:
x (m)
t (s)
J
W0
10
20
30
40
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A student is running at a constant speed of 5 m/s in an attempt to catch a Jammie Shuttle. When she is 11 m from the bus, it pulls away with a constant acceleration of 1 m/s2. From this point, how long does it take her to catch up to the bus if she keeps running with the same speed? Mathematical representation:
(x1)W = (x0)W + (v0x)W t + ½ (ax)W t2
(x1)J = (x0)J + (v0x)J t + ½ (ax)J t2
She catches the shuttle when (x1)W = (x1)J
i.e. 5t = 11 + ½t2
½t2 – 5t + 11 = 0
t = 3.3 s or t = 6.7 s
sf = si + vist + ½as(t)2
= 0 + 5t + (½ 0 t2) = 5t
= 11 + 0t + (½ 1 t2) = 11 + ½t2
NEWTON’S LAWS CONCEPTS OF MOTIONPHY1012F
FREE FALL
The motion of a body moving under the influence of gravity only, and no other forces, is called free fall.
(We often ignore air resistance for slow-moving, massive objects.)
Consequently…
Two objects dropped from the same height in the absence of air resistance will hit the ground simultaneously, at the same speed.
free fallaAny two objects in free fall experience
the same acceleration, .
57
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FREE FALL
Notes: g = 9.80 m/s2 is magnitude of the acceleration due to gravity. It is therefore never negative!
In our convention, afree fall = –g.
g = 9.80 m/s2 is the average value for the surface of the Earth.
“Free fall” refers also to objects which have been projected upwards – not only to those which are literally falling downwards.
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FREE FALL
A kingfisher hovers 30 m directly above a boy with a catapult. If the boy launches a stone straight up at 25 m/s, how long does the stone take to hit the bird?
y
0 y0, v0y, t0
y1, v1y, t1
ay
y0 = 0 m t0 = 0 s
y1 = +30 m
v0y = +25 m/s
ay = –g = –10 m/s2
t1 = ? (= t)
NEWTON’S LAWS CONCEPTS OF MOTIONPHY1012F
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FREE FALL
A kingfisher hovers 30 m directly above a boy with a catapult. If the boy launches a stone straight up at 25 m/s, how long does the stone take to hit the bird?
vy (m/s)
15
25
20
–5
10
0
5
1 2 3t (s)
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61
FREE FALL
A kingfisher hovers 30 m directly above a boy with a catapult. If the boy launches a stone straight up at 25 m/s, how long does the stone take to hit the bird?
y
0 y0, v0y, t0
y1, v1y, t1
ay
y0 = 0 m t0 = 0 s
y1 = 30 m
v0y = 25 m/s
ay = –g = –10 m/s2
t1 = ? (= t)
y1 = y0 + v0yt + ½ay(t)2
30 = 0 + 25t + ½ (–10)t2
5t2 – 25t + 30 = 0
t2 – 5t + 6 = 0
(t – 2)(t – 3) = 0
t = 2 s or t = 3 s ?!
NEWTON’S LAWS CONCEPTS OF MOTIONPHY1012F
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FREE FALL
A kingfisher hovers 30 m directly above a boy with a catapult. If the boy launches a stone straight up at 25 m/s, how long does the stone take to hit the bird?
1v
2v
0v
3v
a
stop/start
5v
4v
NEWTON’S LAWS CONCEPTS OF MOTIONPHY1012F
63
FREE FALL
A kingfisher hovers 30 m directly above a boy with a catapult. If the boy launches a stone straight up at 25 m/s, how long does the stone take to hit the bird?
vy (m/s)
15
25
20
–5
10
0
5
1 2 3t (s)
NEWTON’S LAWS CONCEPTS OF MOTIONPHY1012F
64
Sign chosen by inspection.(In cases where the acceleration points left, as = –g sin )
MOTION ON AN INCLINED PLANE
The acceleration down (i.e. parallel to) this frictionless plane which is inclined at an angle to the horizontal is…
sv
sa
a
a
free fall
a
g
as = g sin
a
v
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INSTANTANEOUS ACCELERATION
For non-uniformly accelerated motion we can define instantaneous acceleration similarly to the way we defined instantaneous velocity…
0lim s s
st
v dva
t dt
Mathematically, as
Graphically, as the slope of the tangent to the velocity curve at a specific instant of time t.
vs (m/s)
t (s)t
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KINEMATICS
Learning outcomes:At the end of this chapter you should be able to…
Interpret, draw and convert between position, velocity and acceleration graphs.
Use an explicit problem-solving strategy for kinematics problems.
Apply appropriate mathematical representations (equations) in order to solve numerical kinematics problems.