RELATIV VELOCITY ON A STRAIGHT ROAD
You drive north on a straight two-lane road at a constant 88 km/h. A truck in the other lane approaches you at a constant 104 km/h (Fig. 3.33). Find (a) the truck’s velocity relative to you and (b) your velocity relative to the truck. (c) How do the relative velocities change after you and the truck pass each other? Treat this as a one-dimensional problem.
IDENTIFY and SET UP:
In this problem about relative velocities along a line, there are three reference frames: you (Y), the truck (T), and the earth’s surface (E). Let the positive x-direction be north (Fig. 3.33). Then your x-velocity relative to the earth is v_{\mathrm{Y} / \mathrm{E}-x}=+88 \mathrm{~km} / \mathrm{h}.
The truck is initially approaching you, so it is moving south and its x-velocity with respect to the earth is v_{\mathrm{T} / \mathrm{E}-x}=-104 \mathrm{~km} / \mathrm{h}.
The target variables in parts (a) and (b) are v_{\mathrm{T} / \mathrm{Y}-x} \text { and } v_{\mathrm{Y} / \mathrm{T}-x}, respectively. We’ll use Eq. (3.32) (v_{P / A-x}=v_{P / B-x}+v_{B / A-x}) to find the first target variable and Eq. (3.33) (v_{A / B-x}=-v_{B / A-x}) to find the second.
EXECUTE:
(a) To find v_{\mathrm{T} / \mathrm{Y}-x} we write Eq. (3.32) (v_{P / A-x}=v_{P / B-x}+v_{B / A-x}) for the known v_{\mathrm{T} / \mathrm{E}-x} and rearrange:
v_{\mathrm{T} / \mathrm{E}-x}=v_{\mathrm{T} / \mathrm{Y}-x}+v_{\mathrm{Y} / \mathrm{E}-x}v_{\mathrm{T} / \mathrm{Y}-x}=v_{\mathrm{T} / \mathrm{E}-x}-v_{\mathrm{Y} / \mathrm{E}-x}
= -104 km/h – 88 km/h = -192 km/h
The truck is moving at 192 km>h in the negative x-direction (south) relative to you.
(b) From Eq. (3.33) (v_{A / B-x}=-v_{B / A-x}),
v_{\mathrm{Y} / \mathrm{T}-x}=-v_{\mathrm{T} / \mathrm{Y}-x}=-(-192 \mathrm{~km} / \mathrm{h})=+192 \mathrm{~km} / \mathrm{h}You are moving at 192 km/h in the positive x-direction (north)
relative to the truck.
(c) The relative velocities do not change after you and the truck pass each other. The relative positions of the bodies don’t matter.
After it passes you the truck is still moving at 192 km/h toward the south relative to you, even though it is now moving away from you instead of toward you.
EVALUATE: To check your answer in part (b), use Eq. (3.32) (v_{P / A-x}=v_{P / B-x}+v_{B / A-x}) directly in the form v_{\mathrm{Y} / \mathrm{T}-x}=v_{\mathrm{Y} / \mathrm{E}-x}+v_{\mathrm{E} / \mathrm{T}-x}. (The x-velocity of the earth with respect to the truck is the opposite of the x-velocity of the truck with respect to the earth: v_{\mathrm{E} / \mathrm{T}-x}=-v_{\mathrm{T} / \mathrm{E}-x}.) Do you get the same result?