If an object moves relative to a reference frame (for example, if a professor moves to the right relative to a white board or a passenger moves toward the rear of an airplane), then the object’s position changes. This change in position is known as displacement. The word “displacement” implies that an object has moved, or has been displaced.

Displacement is the *change in position* of an object:

Δx=xf−x0,Δx=xf−x0, size 12{Δx=x rSub { size 8{f} } - x rSub { size 8{0} } } {}

(1)where ΔxΔx size 12{Δx} {} is displacement, xfxf size 12{x rSub { size 8{f} } } {} is the final position, and x0x0 size 12{x rSub { size 8{0} } } {} is the initial position.

In this text the upper case Greek letter ΔΔ size 12{Δ} {} (delta) always means “change in” whatever quantity follows it; thus, ΔxΔx size 12{Δx} {} means *change in position*. Always solve for displacement by subtracting initial position x0x0 size 12{x rSub { size 8{0} } } {} from final position xfxf size 12{x rSub { size 8{f} } } {}.

Note that the SI unit for displacement is the meter (m) (see Physical Quantities and Units), but sometimes kilometers, miles, feet, and other units of length are used. Keep in mind that when units other than the meter are used in a problem, you may need to convert them into meters to complete the calculation.

Note that displacement has a direction as well as a magnitude. The professor’s displacement is 2.0 m to the right, and the airline passenger’s displacement is 4.0 m toward the rear. In one-dimensional motion, direction can be specified with a plus or minus sign. When you begin a problem, you should select which direction is positive (usually that will be to the right or up, but you are free to select positive as being any direction). The professor’s initial position is x0=1.5mx0=1.5m size 12{x rSub { size 8{0} } =1 "." 5`m} {} and her final position is xf=3.5mxf=3.5m size 12{x rSub { size 8{f} } =3 "." 5`m} {}. Thus her displacement is

Δx=xf−x0=3.5 m−1.5 m =+2.0 m.Δx=xf−x0=3.5 m−1.5 m =+2.0 m. size 12{Δx=x"" lSub { size 8{f} } - x rSub { size 8{0} } =3 "." 5`m - 1 "." 5`"m "= +2 "." "0 m"} {}

(2)In this coordinate system, motion to the right is positive, whereas motion to the left is negative. Similarly, the airplane passenger’s initial position is x0=6.0 mx0=6.0 m and his final position is xf=2.0 mxf=2.0 m size 12{x rSub { size 8{f} } =2 "." 0`m} {}, so his displacement is

Δx=xf−x0=2.0 m−6.0 m=−4.0 m.Δx=xf−x0=2.0 m−6.0 m=−4.0 m. size 12{Δx=x"" lSub { size 8{f} } - x rSub { size 8{0} } =2 "." 0`m - 6 "." 0`m= - 4 "." 0`m} {}

(3)His displacement is negative because his motion is toward the rear of the plane, or in the negative xx size 12{x} {} direction in our coordinate system.

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