Kinematics – describes the motion of object without causes that leaded to the motion

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.

r(t1)r(t2)r(t3)

x

y

t=t1 t=t2 t=t3

x1=19m,t1=1sx2=277m,t1=4s

),,()( 2222 zyxtr

011 zy

022 zy

),,()( 1111 zyxtr

Very often I will write rn instead of r(tn), the same for the components of the vector, for example, yn instead of y(tn)

Motion along a straight line• x component of a displacement vector – 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)

r(t1)r(t2)

x

y

12 rrd

12 '' rrd

r’ 1

r’ 2

x’y’

• So we can define change in time t=t2-t1 – or in other words a time interval during which x occurs.

• We will define x-component of an object’s average velocity = x/t=(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)

t (sec)

Dt

Dx

•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 quantity X (m)

t (sec)

Two time intervals

Motion in positive x direction

Motion in negative x-direction

Distance is equal to sum of both

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).

0

1

2

3

4

5

6

7

8

9

10

0 1 2 3 4 5 6 7

x(m)

t (s)A

B

C

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.

0

1

2

3

4

5

6

0 1 2 3 4 5 6 7 8 9

X(m)

t(s)

x

t

Case we have more curves that describes motion several objects

Points were graphs cross each other – place and time where both meet each other

Instantaneous velocity

12

12

0 12 tt

xxLim

t

xLimv

tttx

•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

•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.

12

12

0 12 tt

xxLim

t

xLimv

ttt

t1 t2

X (m)

t (sec)

We see that instantaneous velocity is slope of a tangent at point t1

X (m)

t (sec)

Positive slope

Slope decreasing – slowing down, since motion in positive x

Slope is zero – instant stop

Slope is increasing – speeding up, since motion in positive x

Zero slope - stopped

Slope is decreasing, motion in negative direction, object speeding up

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=t2-t1, 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 x-component of it.

And can be positive or negative

t1 t2

v (m/s)

t (sec)

Dt

Dvx=vx2-vx1

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

t1 t2

v (m/s)

t (sec)

We see that instantaneous acceleration is slope of a tangent to vx(t) curveat point t1

12

12

0 12 tt

vvLim

t

vLima xx

tt

x

tx

Reducing time interval we can define an instantaneous acceleration

Special cases

Object is not moving:

Object is moving with constant velocity – means slope to x(t) curve is constant – means it is again a straight line:

0)( xtx 0)()( tvtvv xxav

0)( xxxav vtvv

0)( taa xxav

0)( taa xxav

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

tvxtx x 00)(

x

t

x0

x

t

x0

vx

t

vx0

vx

t

x

t

x0

vx

t

vx0

t1 t2

x1

x2

t1 t2

Area of shaded rectangular region is equal to vx0(t2-t1)=vx0Dt =Dx

x2-x1=Dx=vav-xDt=vx0Dt

Area of a region created by vx(t) curve, t=t1, t=t2 and vx=0 is equal to x-component 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

)(2

1)(

2)(

221

22120

201012

2112 ttattv

tavtavtt

sidesidehAreaxx xx

xxxx

200 2

1)( tatvxtx xx

tvvxttavxtx xxxx )(2

1)2(

2

1)( 0000

Expression for x-component of the position vector, when x0, t, v0x, ax are given

Expression for x-components, when x0 ,t, v0x, vx are given (acceleration is not given)

We can set t1=0 and write x1 as x0, indicating that position at t=0

)(2 1221

22 xxavv xxx 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

tgvv yy 0

•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 y-component of the acceleration is equal to –g.

•We can use all equations for constant acceleration with ay=-g.

200 2

1)( tgtvyty y

)(2 1221

22 yygvv yy

hgvyygv yy 2)(20 2112

21

Special case when you throw object upward, how high does it get?

g

vh y

2

21

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 high-acceleration 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?