Relative Velocity

How are observations that are made in different reference frames related to each other? Relative Velocity

What speed does the observer see each of the trains moving at?

What speed does an observer on the express train see the local train moving at?

When the velocities are along the same line, we can just add or subtract to find the relative velocity.

We need to use vector addition to find relative velocity. First, we need to sketch and label the situation. Second, we need to label each velocity with subscripts that identify the object and the reference frame Relative Velocity

Veg=velocity of express train relative to the ground Reference frame In which velocity applies Veg Veg=velocity of express train relative to the ground Veg= 60mph N Vlg=35mph N Note: Vlg = -Vgl object

Relative to express Vle Vle= Vlg + Vge Notice that inner subscripts (rhs) are the same. Outer subscripts (rhs) are the same as subscripts on lhs. Vle= 35mph +(-60mph) Vle=-25mph local

Vle Vle= Vlg + Vge Vle= -35mph +(-60mph) Relative to express local

Vel Vel= Veg + Vgl Vel= 60mph +(35mph) Relative to local express

The engine of boat drives it across a river that is 1800m wide The engine of boat drives it across a river that is 1800m wide. The velocity of the boat relative to the water is 4.0m/s, directed perpendicular to the current. The velocity of the water relative to the shore is 2.0m/s. A) What is the velocity of the boat relative to the shore? Vbw = velocity of the boat relative to the water Vws = velocity of the water relative to the shore Find vbs = velocity of boat relative to shore

vbs vbw vws A) What is the velocity of the boat relative to the shore? Vbw = 4.0 m/s Vws = 2.0 m/s vbs = vbw + vws = 4.5 m/s (using the pythagorean theorem) tanƟ = vbw / vws so Ɵ=63°

vab=vag+vgb Vag=25m/s Vbg=15.8m/s Two cars are approaching an intersection along perpendicular roads. The cars have velocities as shown in the figure. A) What is the velocity of car A as measured by a passenger in car B. vab=vag+vgb Vag=25m/s A Vbg=15.8m/s B

Homework: p. 84 #54, 55, 56, 57, 60, 61