Foundation course - PHYSICS
Lecture 5-1: Kinematics of a particle
● motion in two dimensions
● angled launches
● uniform and nonuniform circular motion
● relative motion
Naďa Špačková spackova@physics.muni.cz
Summary
v = v0 + at x=x0+v0 t+
1
2
at
2
Motion with a constant acceleration
Time
Position
Velocity
vi
xi
a
Δv
v
Free fall
Free fall is the motion of an object when gravity is the only significant force
acting on it.
Galileo’s experiments with
free fall motion
g
ay =−g
vy=−gt
y = h −
1
2
gt2
y
x
h
v0y
= 0
Free fall
Free-fall acceleration:
● near Earth’s surface is about 9.8 m/s2
downward (each second
velocity increases by 9.8 m/s)
● it is not dependent on mass, density and shape of the falling object
● its magnitude depends on distance from the Earth
Without air friction (in vacuum) all objects
are falling equally.
Downward throw
g
y
x
h
v0y
≠ 0
v0y
vy =−v0 y − gt
y = h − v0 y t −
1
2
gt2
Downward throw
● initial velocity is not zero and is in the same direction as free fall acceleration
Upward throw
Upward throw
● initial velocity is not zero and is in the opposite direction as free fall acceleration
g
x
ymax
v0y
≠ 0
v0y
vy = v0 y − gt
y = y0 + v0 y t −
1
2
gt2
Motion in two dimensions
red ball = no initial velocity
blue ball = initial horizontal velocity
balls have the same vertical motion as they fall
Independence of motion in two
dimensions
● motion of a projectile (blue ball) is a
combination of two motions – horizontal
and vertical
Horizontal motion does not affect
vertical motion and vice versa
Downward velocity increases regularly due
to free-fall acceleration, horizontal velocity
(blue ball) is constant
Angled launches
When a projectile is launched at an angle, the initial velocity has
a vertical component as well as horizontal component.
Angled launches
v0 x = v0 cosθ v0 y = v0 sinθ
v0x
v0y
v0
θ
Separation of vertical and horizontal motions:
● horizontal motion → horizontal velocity component v0x
● vertical motion → vertical velocity component v0y
We know: We want to know:
magnitude v0
and direction θ v0x
and v0y
components
v0x
and v0y
components magnitude v0
and direction θ
v0 = √v0 x
2
+ v0 y
2
tanθ =
v0 y
v0 x
Horizontally launched projectile
The horizontal and vertical velocity
components at each moment are
added to form the velocity vector at
that moment.
The trajectory has a parabolic
shape.
x = v0 t + x0
y = h −
1
2
gt
2
horizontal motion: constant velocity
vertical motion: constant acceleration
h
Example: horizontally launched projectile
A projectile is fired horizontally from a gun that is 45.0 m above flat ground, emerging from
the gun with a speed of 250 m/s.
(a) How long does the projectile remain in the air?
(b) At what horizontal distance from the firing point does it strike the ground?
(c) What is the magnitude of the vertical component of its velocity as it strikes the ground?
Angled launches
When a projectile is launched at an angle,
the initial velocity has a vertical component
as well as horizontal component.
Vertical upward and
downward motion:
● at each point the velocity of
the object has the same
magnitude
● the directions of the velocities
are opposite
horizontal motion: constant velocity
vertical motion: constant acceleration
Angled launches
Maximum height:
● vertical velocity vy
is zero
x − x0 = v0 x t
y − y0 = v0 y t −
1
2
gt
2 v0 x = v0 cosθ
v0 y = v0 sinθ
v y = v0 y − gt
h =
v0
2
sin
2
θ
2 g
A
Angled launches
Range:
● horizontal distance when initial and final
heights are the same ( y – y0
is zero )
x − x0 = v0 x t
y − y0 = v0 y t −
1
2
gt
2
v0 x = v0 cosθ
v0 y = v0 sinθ
vy = v0 y − gt
R =
v0
2
g
sin 2θ
B
Angled launches
A projectile fired from the origin with
a given speed at various angles of
projection.
Complementary values of the angle
result in the same value of x (range
of the projectile).
For a given v0
the maximal range is
obtained with θ = 45°.
Range-angle dependency:
Checkpoint question:
Figure shows three paths for a football kicked
from ground level. Ignoring the effects of air, rank
the paths according to:
(a) time of flight
(b) initial vertical velocity component
(c) initial horizontal velocity component
(d) initial speed
Example: Cannonball to pirate ship
Figure shows a pirate ship 560 m from
a fort defending a harbor entrance. A
defense cannon, located at sea level,
fires balls at initial speed v0
= 82 m/s.
At what angle θ0
from the horizontal
must a ball be fired to hit the ship?
Uniform circular motion
It is a motion with acceleration related to the change in velocity
direction.
Uniform circular motion
● movement of an object at a constant speed around a circle with a fixed radius
● position of the object is given by the position vector r
r1
r2
r1
v1
v2
Δr
Position vectors, velocity vectors,
displacement vector
Object’s velocity is defined as:
¯v =
Δ r
Δt
¯v =
Δ x
Δt
In case of circular motion:
Velocity vector is tangent to the circular path.
r1
r2
r1
v1
v2
v1
v2
Δv
a
a
¯a =
Δ v
Δt
The average acceleration:
Centripetal acceleration a:
Δr
r
=
Δ v
v
Δr
r Δt
=
Δ v
v Δt
v
r
=
a
v a=
v2
r
Uniform circular motion
The average acceleration has the same
direction as Δv.
For a very small time interval, a points toward
the center of the circle.
Period of revolution T:
● time needed for the object to make one complete revolution
● during this time the object travels a distance equal to the circumference
of the circle (2πr)
T=
2πr
v
Uniform circular motion
Analogical to the equation for the uniform motion with constant
velocity s = v t
v =
2πr
T
2π
T
= ω
Angular velocity ω
v = ωr
Angular velocity is measured in radians (dimensionless unit). Radian
is related to the ratio of the circumference of a circle to its radius.
f =
1
T
Frequency f:
Frequency units: 1 Hz (Hertz) = 1 s-1
Nonuniform circular motion
Velocity vector changes not only its direction, but also its magnitude.
Acceleration is the sum of its radial and tangential components:
● radial component ar
arises from the change in direction of the velocity
● vector and is directed toward the center of curvature
● tangential component at
causes the change in magnitude of the velocity
vector (at
becomes zero, if the particle follows uniform circular motion)
unit vectors θ and r
determine radial and
tangential direction
a = ar
+ at
Checkpoint questions:
(a) Is it possible to be accelerating while traveling at constant speed?
(b) Is it possible to round a curve with zero acceleration?
(c) Is it possible to round a curve with a constant magnitude of acceleration?
Figure shows four tracks (either half- or
quarter-circles) that can be taken by a train,
which moves at a constant speed.
Rank the tracks according to the magnitude
of a train’s acceleration on the curved
portion, greatest first.
Example: uniform circular motion
An Earth satellite moves in a circular orbit 640 km above Earth’s surface with a period
of 98.0 min. What are the
(a) speed,
(b) magnitude of the centripetal acceleration of the satellite?
Earth’s radius is 6371 km.
Relative motion in one dimension
vbus/ground
vperson/bus
observer staying
on the street
relative to bus
vperson/ground
relative to observer
on the street
vperson/bus
relative to bus
vperson/ground
relative to observer
on the street
observer staying
on the street
vbus/ground
Velocity of an object depends on the reference frame where it is
observed or measured.
Relative motion in one dimension
P
BA
vPB
vBA
xBA
xPB
xPA
= xPB
+ xBA
Position (displacement) of the particle:
xPA
= xPB
+ xBA
Velocity of the particle:
vPA
= vPB
+ vBA
Reference frames A and B move at
constant velocity relative to each other.
Acceleration of the particle:
vBA
is constant → aBA
is zero
aPA
= aPB
Observers on different frames of reference that move at constant velocity
relative to each other will measure the same acceleration for a moving
particle.
Example: In figure above suppose that Barbara’s velocity relative to Alex
is constant vBA
= 52 km/h and car P is moving in the negative
direction of the x axis.
(a) If Alex measures a constant vPA
= -78 km/h for car P, what
velocity vPB
will Barbara measure?
(b) If car P brakes to a stop relative to Alex in time t = 10 s at
constant acceleration, what is its acceleration aPA
relative to
Alex?
(c) What is the acceleration aPB
of car P relative to Barbara
during the braking?
P
BA
vPB
vBA
A… observer Alex
B… observer Barbara
P… car
Relative motion in two dimensions
x
y x’
y’
vBA
vPB
vPA
α
γ
β
Combining velocities:
● resolve the vectors into x- and
y-components
vBA
:
x-component: vBAx
= vBA
cos α
y-component: vBAy
= vBA
sin α
vPB
:
x-component: vPBx
= vPB
cos β
y-component: vPBy
= vPB
sin β
vPA
:
x-component: vPAx
= vPB
cos β + vBA
cos α
y-component: vPAy
= vPB
sin β + vBA
sin α
vPB + vBA = vPA
P
B
A
vPA
2
= vPAx
2
+ vPAy
2
Example: relative motion in two dimensions
A train travels due south at 30 m/s (relative to the ground) in a rain that is blown toward
the south by the wind. The path of each raindrop makes an angle of 70° with the
vertical, as measured by an observer stationary on the ground. An observer on the
train, however, sees the drops fall perfectly vertically. Determine the speed of the
raindrops relative to the ground.