Relative Velocity

Subscript conventionVab = Velocity of ‘a’ as measured from ‘b’ Or

Velocity of ‘a’ relative to ‘b’

adb = Acceleration of ‘d’ as measured from ‘b’ OrAcceleration of ‘d’ relative to ‘b’

One Dimensional Relative motion

Let A and B have constant relative velocity. But P can change it’s velocity

How do the subscripts work?

In the Figure , suppose that Barbara's velocity relative to Alex is a constant VBA = 52 Km/hr and car P is moving in the negative direction of the x axis

(a) If Alex measures a constant for car P, VPA = -78 Km/hr, what velocity will Barbara measure?

(b) If car P brakes to a stop relative to Alex (and thus relative to the ground) in time at constant acceleration, what is its acceleration relative to Alex?

The initial velocity of P relative to Alex is -78 Km/hr and the final velocity is 0. Thus, the acceleration relative to Alex is

(c) What is the acceleration of car P relative to Barbara during the braking?

We know the initial velocity of P relative to Barbara from part (a) is VPB = -130 Km/ hr. The final velocity of P relative to Barbara is -52 km/h (this is the velocity of the stopped car relative to the moving Barbara). Thus,

3.3 Relative Velocity

A student walks on a treadmill moving at 4.0 m/s and remains at the same place in the gym.

a)What is the student�s velocity relative to the gym floor?

b) What is the student�s speed relative to the treadmill?

c) What is the treadmill�s speed relative to the student?

s = student, t = treadmill, and g = groundUnknown: vst. Knowns: vsg = 0, vtg = -4.0 m/s.

vsg = vst + vtg → vst = vsg – vtg = 0 – (-4.0 m/s) = 4.0 m/s

0

vts = - vst = -4.0 m/s

Two Dimensional Relative motion

Our two observers are again watching a moving particle P from the origins of reference frames A and B, while B moves at a constant velocity VBA relative to A.

Example: A river has a current with a velocity of 1.0 m/s south. A boat, whose speed in still water is 5.0 m/s, is directed east across the 100 m wide river.

a) How long does it take the boat to reach the opposite shore?b) How far downstream will the boat land?c) What is the velocity of the boat relative to the shore?

Ө

vbr = 5.0 m/s

vbs

x

y

vrs = 1.0 m/s

Solution:Use the subscripts: r = river, b = boat, s = shore

Given: vrs = 1.0 m/s vbr = 5.0 m/s x = 100 m

Unknowns: a) t b) y c) vbs

a) Use kinematics equation 2: x = xo + vot +1/2 at2 , where a = 0, xo = 0, vo = vbr, and x = 100 m

x= vot → solve for t t = x = 100 m = 20 svo 5.0 m/s

b) Use kinematics equation 2: y = yo + vot +1/2 at2 , where a = 0, yo = 0, vo = vrs

y = vot = (1.0 m/s)(20 s) = 20 m

c) The velocity of the boat relative to shore is the vector sum of the velocity of the boat relative to the river and the velocity of the river relative to the shore (current) vbs = vbr + vrs

*Hint: the pattern of subscripts are helpful in problem solving. On the right side of the equation, the two inner subscripts are the same (r). The outer subscripts (b and s) are sequentially the same as those for the relative velocity on the left side of the equation. ***

vbs = √vbr2+ vrs2 = √(5.0 m/s)2 + (1.0 m/s)2 = 5.1 m/s

Ө = tan-1 5.0 m/s = 79�measured from the shoreline.1.0 m/s

Example: If the person on the boat in the previous example wants to travel directly across river,a)What angle upstream must the boat be directed?b)With what speed will the boat cross the river?c)How long will it take the boat to reach the opposite shore?

Solution:To travel directly across the river, the velocity of the boat relative to the shore must be directly across the river. The vector form of the relative velocity equation

vbs = vbr + vrs is still valid.

a)Ө = sin -1 vrs = sin-1 1.0 m/s = 12�vbr 5.0 m/s

b) From the triangle in the diagram,vbs2 + vrs 2 = vbr 2 so,

vbs = √ vbr 2 - vrs 2

vbs = √ (5.0 m/s)2 – (1.0 m/s)2 = 4.9 m/s

c) The time is then t = y = 100 m = 20 svbs 4.9 m/s

Ө

vbr = 5.0 m/s

vbs

x

yvrs = 1.0 m/s

Circular MotionnUniform circular motion (constant centripetal

acceleration)

nMotion with a tangential and radial (centripetal)acceleration

Uniform Circular Motion• Object has a circular path, constant

speed.

• The velocity is always tangent to the path of the object.

Why is there acceleration in uniform circular motion?

Dv = vf – vi

a= Dv/ Dt

Centripetal Acceleration• Vector

• Always perpendicular to the path of the motion.

• Points toward the center of the circle.

2

Cvar

=

Characteristics of UniformCircularMotion

• Tangential (linear) Velocity• Frequency• Period• Centripetal Acceleration

Frequency, f :#revolutions per unit time

• f = # rev / time

Units:• (1/sec)=sec-1=Hertz (Hz)• rpm (#rev/min)• rps (#rev/sec)

r

Period

• Period T : time for 1 revolution–Unit: sec, min, h

• Relating Frequency and periodf= 1

T

Tangential Speed in terms of T or f

v=2pr/T

v=2prf

Prove of a = v2/rGo over the book

Example 1

• A boy whirls a stone in a horizontal circle of r=1.5m, 2m above the ground. The string breaks and the stone strikes 10m away. What was the centripetal acceleration during the circular motion?

• Answer: 160m/s2

When Acceleration is not uniform -Tangential Acceleration exists

n When v varies.

n Motion is not uniform circular motion.

Total Acceleration• Tangential

acceleration (vector)

Plus

• Radial acceleration (vector)

Total Acceleration

• Tangential acceleration:

• Radial acceleration:

• Total acceleration:

td

adt

=v

2

r Cva ar

= - = -

2 2r ta a a= +

Total Acceleration

– r, radius vector- q , tangent to the

circle

2ˆ ˆt r

d vdt r

q= + = -v

a a a r

Example 2An automobile whose speed isincreasing at a rate of 0.6m/s2 travels ona circular road of radius 20m. When theinstantaneous speed of the car is 4m/sfind: a)the tangential acceleration, b)thecentripetal acceleration, c)the magnitudeand direction of the total acceleration.

Ans: a)0.6m/s2,b)0.8m/s2, c)1m/s2,53.1o