Kinematics – describes the motion of object without causes that leaded
to the motion
We are not interested in details of the object (it can be car, person, box
etc..). We treat it as a point
We want to describe position of the object with respect to time – we
want to know position at any given time
Path (trajectory) – imaginary line
along which the object moves
x
t
Motion along a straight line
• We will always try to set up our reference frame in a such way
that motion is along or “x” or “y” coordinate axis. The direction
of the axis is up to us.
• Position vector then can be represented by a single component,
the other components are equal to zero.
y
t=t1
t=t2
Very often I will write rn instead of
r(tn), the same for the components of
the vector, for example, yn instead of
t=t3y(t )
n
x
r(t1)
r(t2)
r(t3)
x1=19m,t1=1s
x2=277m,t1=4s

r ( t1 )  ( x1 , y 1 , z 1 )

r (t 2 )  ( x 2 , y 2 , z 2 )
y1  z1  0
y2  z2  0
Motion along a straight
line

 
• x component of a displacement vector d  r2  r1 – for time interval
t1 t2 is equal Dx=x2-x1
• Note that we can define another reference frame, position vector will
be different in each frame, not a displacement vector (it will have
different components)
 

d  r ' 2  r '1
y
x
r(t1)
r(t2)
• So we can define change in time Dt=t2-t1 – or in other words a
time interval during which Dx occurs.
• We will define x-component of an object’s average velocity =
Dx/Dt=(x2-x1)/(t2-t1)=vav. X
• Units - m/s
• X-component of the velocity is equal to a slope of a line in x-t
plane that passes through points (x1,t1) and (x2,t2)
X (m)
Dt
t (sec)
•The value of x-component of an average velocity is defined for
given interval.
•Note, since x-component of the displacement can be positive
(motion in +x) or negative (motion in -x), x-component of an
average velocity can be positive or negative.
•Note that if you start at x1=x2 then x-component of average
velocity for given time interval is equal to zero.
•Average speed=distance traveled / time interval – positive
X (m)
quantity
Two time intervals
Motion in positive x direction
Motion in negative x-direction
Distance is equal to sum of both
t (sec)
Example:
Figure shows the position of a moving object as a function of
time. (a) find the average velocity of this object from points A to
B, B to C and A to C. (b) is the average speed for intervals given in
(a) will be less than, equal, or greater than the values found in
part (a).
x(m)
10
9
C
8
7
6
5
4
3
B
2
1
0
A
0
1
2
3
4
5
6
t (s)
7
Example:
Each graph in figure shows the position of a running cat called Mousie, as
a function of time. In each case, sketch a clear qualitative (no numbers)
graph of Mousie’s velocity as a function of time.
X(m)
6
5
4
3
2
1
0
0
1
2
3
4
5
6
7
8
t(s)
9
x
Case
we have more curves that describes motion several objects
t
Points were graphs cross each other
– place and time where both meet
each other
Instantaneous velocity
•Average velocity as everything that is average does not give us
chance to find out how was the object moving at any given time of
the time interval.
•Instead we define (x-component) an instantaneous velocity,
decreasing time interval.
Dx
x x
v x  Lim
Dt  0
Dt
 Lim
t 2  t1
2
1
t 2  t1
•The same way we can define an instantaneous speed.
•Note: we can always make the time interval small enough so
distance traveled during Dt is equal to |Dx| (motion along straight
line).
•Instantaneous speed is a magnitude of instantaneous velocity
v  Lim
Dt  0
Dx
Dt
 Lim
t 2  t1
x 2  x1
t 2  t1
X (m)
We see that instantaneous velocity is
slope of a tangent at point t1
t1
t2
t (sec)
X (m)
Zero slope - stopped
Slope is increasing –
speeding up, since motion
in positive x
Slope is decreasing, motion in
negative direction, object
speeding up
Slope is zero – instant stop
Slope decreasing – slowing down,
since motion in positive x
Positive slope
t (sec)
Average acceleration
•Suppose that I have a particle that was moving somehow and I
know that x-component of instantaneous velocity is given as vx(t).
•We want to know how much it changed during time interval Dt=t2t1, Dv=vx2-vx1
•X-component of an average acceleration of the particle is equal
to
aav x=Dv/Dt
UNITS:
meters/sec2=m/s2
We are getting average acceleration from instantaneous velocity,
remember it is a vector. In our case we are considering only xcomponent of it.
And can be positive or negative
v
(m/s)
x-component of average acceleration in time interval Dt is a
slope of the straight line that passes through 2 points of vx vs t
curve
Dvx=vx2-vx1
Dt
t1
t2
t (sec)
v
(m/s)
We see that instantaneous acceleration is slope of a
tangent to vx(t) curveat point t1
Reducing time interval we can define an
instantaneous acceleration
a x  Lim
Dt  0
t1
Dv x
Dt
 Lim
t 2  t1
t2
v x 2  v x1
t 2  t1
t (sec)
Special cases
Object is not moving:
x (t )  x 0
v av  x  v x ( t )  v ( t )  0
vx
x
x0
a av  x  a x ( t )  0
t
t
Object is moving with constant velocity – means slope to x(t)
curve is constant – means it is again a straight line:
x (t )  x 0  v x 0  t
v av  x  v x ( t )  v x 0
a av  x  a x ( t )  0
x
vx
vx0
x0
t
t
Motion with constant velocity differs from motion with constant
speed. The words constant velocity locks magnitude and
direction – means object is moving along straight line
x
x2
x2-x1=Dx=vav-xDt=vx0Dt
x1
x0
vx
vx0
t1
t2
t
t1
t2 t
Area of shaded rectangular region is
equal to vx0(t2-t1)=vx0Dt =Dx
Area of a region created by vx(t) curve, t=t1, t=t2 and vx=0 is equal to xcomponent of displacement.
Motion with constant acceleration
•
•
•
•
Instantaneous and average acceleration are equal
x-component of velocity is given: vx(t)=v0x+ax t
One can see that at time t=0 vx=v0x
We know that displacement is equal to area bound by velocity versus time, t=t1,
t=t2 and vx=0 lines – trapezoid.
• Height =h= t2-t1 two parallel sides: side1=v0x+a t1 and side2=v0x+a t2
x 2  x1  Area  h
side 1  side
2
2
 ( t 2  t1 )
v 0 x  a x t1  v 0 x  a x t 2
2
 v 0 x ( t 2  t1 ) 
1
2
a x ( t 2  t1 )
2
2
We can set t1=0 and write x1 as x0, indicating that position at t=0
x (t )  x 0  v 0 x t 
x (t )  x 0 
1
2
1
2
a xt
2
Expression for x-component of the position
vector, when x0, t, v0x, ax are given
(2v0 x  a xt )  t  x0 
1
2
(v0 x  v x )  t
Expression for x-components,
when x0 ,t, v0x, vx are given
(acceleration is not given)
2
2
v 2 x  v1 x  2 a x ( x 2  x1 ) Expression for x-components, when x2 –x1, v1x,
v2x,,ax are given. (time is not given)
Example: According to recent data, a Ford Focus travels 0.25mi in
19.9s, starting from rest. The same car, when braking from 60mph
on dry pavement, stops in 146ft. (a) Find this car’s acceleration
and deceleration, (b) assuming constant acceleration find its
velocity after 0.25mi.
Freely falling objects
•In this section we will consider constant acceleration due to gravity.
•Magnitude of the acceleration we will denote as g=9.8m/s2 (for Earth)
•Acceleration due to gravity is a vector pointing toward the center of the
(in our case Earth)
•we can choose our reference frame with y-axis pointing up – then ycomponent of the acceleration is equal to –g.
•We can use all equations for constant acceleration with ay=-g.
v y  v0 y  g  t
y (t )  y 0  v 0 y t 
1
Special case when you throw object
upward, how high does it get?
g t
2
2
0  v1 y  2 g  ( y 2  y 1 )  v1 y  2 g  h
2
2
v 2 y  v1 y  2 g  ( y 2  y 1 )
2
2
2
h
v1 y
2g
Example:
If we throw a ball upward with initial velocity v0y=5m/s, find:
• how high does it go,
• how long does it take to get there
• velocity of the ball when it is on its way back
• time required to return back
Example: Measuring depth of a deep well
To measure the depth of the well you drop a rock and start your
stopwatch. Find the depth of the well if after 3s you hear hitting
sound coming from a bottom of the well.
Ignore finite speed of a sound.
Example:
Two rockets start from rest and accelerate to the same final speed,
but one has twice the acceleration of the other. (a) If the highacceleration rocket takes 50s to reach the final speed, how long will it
take the other rocket to reach that speed. (b) If the high acceleration
rocket travels 250m to reach its final speed, how far will the other
rocket travel to do likewise?