RELATIVE MOTION ANALYSIS (Section 12.10) Objectives: a)Understand translating frames of reference. b)Use translating frames of reference to analyze relative motion.

APPLICATIONS The boy on the ground is at d = 3m when the ball was thrown to him from the window. If the boy is running at a constant speed of 1.2m/s, how fast should the ball be thrown? If aircraft carrier travels at 50km/hr and plane A takes off at 200 km/hr (in reference to water), how do we find the velocity of plane A relative to the carrier? the same for B?

RELATIVE POSITION The absolute position of two particles A and B with respect to the fixed x, y, z reference frame are given by r A and r B. The position of B relative to A is represented by r B/A = r B – r A Therefore, if r B = (10 i + 2 j ) m andr A = (4 i + 5 j ) m, thenr B/A = (6 i – 3 j ) m.

RELATIVE VELOCITY The relative velocity of B with respect to A is the time derivative of the relative position equation: v B/A = v B – v A or v B = v A + v B/A v B and v A are absolute velocities and v B/A is the relative velocity of B with respect to A. Note that v B/A = - v A/B.

RELATIVE ACCELERATION The time derivative of the relative velocity equation yields a similar vector relationship between the absolute and relative accelerations of particles A and B. a B/A = a B – a A or a B = a A + a B/A

Solving Problems Two ways of solution: 1. The velocity vectors v B = v A + v B/A could be written as Cartesian vectors and the resulting scalar equations solved for up to two unknowns. 2. Or can be solved “graphically” by use of trigonometry. This approach is not recommended.

GRAPHICAL SOLUTION LAWS OF SINES AND COSINES a b c C B A Since vector addition or subtraction forms a triangle, sine and cosine laws can be applied to solve for relative or absolute velocities and accelerations. Law of Sines: C c B b A a sin  Law of Cosines: Abccb a cos    Bac c a b cos   C ab b a c cos2 222 

EXAMPLE Given: v A = 600 km/hr v B = 700 km/hr Find: v B/A

EXAMPLE (continued) v B/A = v B – v A = ( i j ) km/hr hr km v A B ) ( ) ( 22 /    where     ) ( tan 1   Solution: v A = 600 cos 35 i – 600 sin 35 j = (491.5 i – j ) km/hr v B = -700 i km/hr a) Vector Method:

EXAMPLE (continued) Law of Sines:  sin)145°sin( / A AB v v  or    b) Graphical Method:  145 v B = 700 km/hr v A = 600 km/hr v B/A Note that the vector that measures the tip of B relative to A is v B/A. Law of Cosines:   145cos ) 600 )( 700 ( 2 ) 600 () 700 ( / A B v  hr km v AB / 

GROUP PROBLEM SOLVING Given: v A = 10 m/s v B = 18.5 m/s a t ) A = 5 m/s 2 a B = 2 m/s 2 Find: v A/B a A/B Solution: The velocity of Car A is: v A = y x

GROUP PROBLEM SOLVING (continued) The velocity of B is: v B = The relative velocity of A with respect to B is (v A/B ): v A/B =

GROUP PROBLEM SOLVING (continued) a B = 2i (m/s 2 ) a A = (a t ) A + (a n ) A =[5 cos(45)i – 5 sin(45)j] + [-( ) sin(45)i – ( ) cos(45)j] a A = 2.83i – 4.24j (m/s 2 ) The relative acceleration of A with respect to B is: a A/B = a A – a B = (2.83i – 4.24j) – (2i) = 0.83i – 4.24j a A/B = (0.83) 2 + (4.24) 2 = 4.32 m/s 2  = tan -1 ( ) = 78.9° 