Motion in a Plane• We set up a reference frame.• Now position of the object is defined by a position vector r that may have

non-zero components (x,y)• Suppose at t=t1 it is r1 (x1,y1) and t=t2 r2 (x2,y2)• We can define a displacement vector: d= r2 -r1=r• And a vector of average velocity:

t

y

tt

yyvavy

12

12

t

x

tt

xxvavx

12

12

t

r

tt

d

tt

rrvav

1212

12 Average velocity parallel to the displacement vector

tvx xav .

xav

yav

v

v

x

yTan

.

.

tvy yav .

An imaginary line along which an object was moving-path

Note: Velocity vector is along the displacement vector

Instantaneous velocity

t

rLimv

t

0 t

xLimv

tx

0 t

yLimv

ty

0x, y components

•What happens if one decreases time interval Dt ? Both, Dx Dy, x, y- components of the displacement vector will get smaller… And we can see that the instantaneous velocity will be tangent to the path.

•Magnitude of the instantaneous velocity is called instantaneous speed v=(vx

2+vy2)1/2.

•Example: x, y- components of a position vector that describes motion of a ball are given by equations: x(t)=x0+vx*t and y(t)=y0+vy*t+1/2*ay*t2. Let plot the path of the ball (assume that

ay=-10m/s2. Let vx=1m/s and vy=2m/s, and x0=y0=0m. Find an average velocity between t=1s and t=3s. What its instantaneous velocity at t=1sec (hint: one dimensional motion)

22yx vvv

Average and instantaneous acceleration• Again we measure an instantaneous velocity at time t=t1, v1 and

t=t2=t1+t, v2 .• Average acceleration:

aav=(v2-v1)/(t2-t1)=v/tInstantaneous acceleration:

t

vLima

t

0 t

vLima x

tx

0 t

vLima y

ty

0x, y components

• Note that vector of acceleration – does not need to be along the velocity vector:

• Simple cases:• If it is along – object or speeding up or slowing down• If it is perpendicular – object is turning

• If an object is moving along curved path with constant speed it still have nonzero acceleration – otherwise the object has to move along straight line.

• In case of uniform (constant speed) circular motion, acceleration is along the radius and its magnitude is given as:

22yx aaa

R

varad

2

Centripetal acceleration

Example: A racecar is making loops at constant speed in a circular racetrack. The diameter of the track is 200m, find

an instantaneous velocity of the car at t=0s, 1s, 2s, assuming that it takes 8s to complete one loop.

See what would be average acceleration of the car.

the instantaneous acceleration of the car and its direction

Projectile motion• Motion of an object under gravitational force and air resistance• Important part of the problem – we can treat motion in x and y direction

separately, only time is a common variable.• What does it mean? Object dropped from certain height and thrown

horizontally reach the ground simultaneously. • Setting up reference frame with y-axis perpendicular to the surface, vector of

acceleration has zero x-component and –g y-component.

2/8.9

0

smga

a

y

x

In other words: constant velocity along x-axis and constant acceleration along y-axis

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x

vv

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00

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yygvv

tgvv

tgtvyy

yy

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General setup• Make a simple sketch of the problem and set up reference frame• Initial position: t=0, point from where object was launched (kicked,

thrown, shot, hit) (x0,y0), initial velocity is given using x, y-components v0x,v0y or magnitude v0 (speed) and angle with x-axis or speed and initial trajectory

• Top point: • t2=v0y/g, • x2=x0+v0x(v0y/g), y2=y0+2v2

0y/g, • vx2=v0x , vy2=0• Final point: some problems ask velocity of the projectile when it hits the

ground –pay attention to the sign of y-component of the velocity. • x3=x0+v0xt3 , y3=y0+v0yt3-1/2gt3

2 • vy3=v0y-gt3, vx3=v0x

• Some problems may give information about some additional points along the trajectory of the object. You have to set up time and equations for those points too.

v

Velocity is defined by speed of the bullet and angle that barrel of the rifle makes with horizon. Velocity is tangent to the path – before leaving the barrel bullet moves straight path

v 0

When cargo is just released it has the same velocity the helicopter

Example 3.49: An errand of mercy. An airplane is dropping bales of hay to cattle stranded in a blizzard on the Great Plains. The pilot releases the bales at 150m above the level ground when the plane is flying at 75m/s 55o above the horizontal. How far in front of the cattle should the pilot release the hay so that the bales will land at the point where the cattle are stranded?

When cargo is just released it has the same velocity the helicopter

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00

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vtv

tvxx

x

xx

x

yy

yyy

yy

ata

tavtv

tatvyy

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)(2

1

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200

t

x(t)

t

y(t)

x0

t1

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t1t2

20tv x

y0

10tv y

212

1tay

t1

yy

yy

yt

ata

tatv

tayy

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)(2

1 2

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ta

tv

xx

x

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t

xt

yt

Relative velocity• Suppose we have two students, one at a station and another on a train • Let fix reference frames

xA

yA

zA

xB

yB

zB

rB/A rp/A

rp/B

ABBpAp rrr ///

Why do we need 2 reference frames?

For person on the train table will not move…

For person at the station it can move with train…

Drp/B

Drp/A

rp1/A

rp2/A

rB1/A

rp2/B

rp1/B

rB2

DrB/A

t

r

t

r

t

rABBpAp

///

ABBpAp rrr ///

ABBpAp vvv ///

p/A ->p/B B/A

Example: 3.40. A river flows due to south with a speed of 2m/s. A man steers a motorboat a cross the river; his velocity relative to the water is 4.2m/s due to east. The river is 800m wide.

a) What is his velocity (magnitude and direction) relative to the Earth

b) How much time is required to cross the river?

c) How far south of his starting point will he reach the opposite bank?

Example: 3.42. When a train velocity is 12m/s eastward, raindrops that are falling vertically with respect to the earth make traces that are inclined at 30o to the vertical on the window of the train.

(a)What is the horizontal component of a drops velocity with respect to the earth? With respect to the train?

(b)What is the magnitude of the velocity of the raindrop with respect to the earth? With respect to the train?