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Worked Examples · Example 9.4

Q.Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of R=2 mR = 2\ \text{m}. If the jogger is running at a speed of 5 m s−15\ \text{m s}^{-1}, how fast the image of the jogger appear to move when the jogger is

(a) 39 m39\ \text{m},
(b) 29 m29\ \text{m},
(c) 19 m19\ \text{m}, and
(d) 9 m9\ \text{m} away.
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A car's side-view mirror is convex with f=R/2=+1 mf=R/2=+1\ \text{m}. Differentiating the mirror equation with respect to time gives image speed =(fu−f)2vobject=\left(\frac{f}{u-f}\right)^2 v_{object}. For a jogger approaching at 5 m s−15\ \text{m s}^{-1}, the image speeds are (a) 1320 m s−1≈3.1×10−3 m s−1\frac{1}{320}\ \text{m s}^{-1}\approx3.1\times10^{-3}\ \text{m s}^{-1},

(b) 1180 m s−1≈5.6×10−3 m s−1\frac{1}{180}\ \text{m s}^{-1}\approx5.6\times10^{-3}\ \text{m s}^{-1},

(c) 180 m s−1=1.25×10−2 m s−1\frac{1}{80}\ \text{m s}^{-1}=1.25\times10^{-2}\ \text{m s}^{-1},

(d) 120 m s−1=5.0×10−2 m s−1\frac{1}{20}\ \text{m s}^{-1}=5.0\times10^{-2}\ \text{m s}^{-1} — the image speeds up sharply as the jogger gets closer.

Setting up

A side-view mirror is convex, R=2 mR=2\ \text{m}, so f=R/2=+1 mf=R/2=+1\ \text{m} (positive by the sign convention, since a convex mirror's focus lies behind it).

Using the Cartesian sign convention, an object a distance dd in front of the mirror has u=−du=-d. As the jogger approaches, dd decreases, so uu (which is negative) is becoming less negative — i.e. uu is increasing with time:

dudt=+5 m s−1\frac{du}{dt} = +5\ \text{m s}^{-1}

(The jogger's own speed is 5 m s−15\ \text{m s}^{-1} toward the mirror, which is exactly a rate of increase of uu.)

Differentiating the mirror equation

Start from

1v+1u=1f\frac{1}{v}+\frac{1}{u}=\frac{1}{f}

Differentiate with respect to time (with ff constant):

−1v2dvdt−1u2dudt=0⇒dvdt=−v2u2dudt-\frac{1}{v^2}\frac{dv}{dt} - \frac{1}{u^2}\frac{du}{dt} = 0 \quad\Rightarrow\quad \frac{dv}{dt} = -\frac{v^2}{u^2}\frac{du}{dt}

Since v=fuu−fv=\frac{fu}{u-f}, we have vu=fu−f\frac{v}{u}=\frac{f}{u-f}, so

∣dvdt∣=(fu−f)2∣dudt∣\left|\frac{dv}{dt}\right| = \left(\frac{f}{u-f}\right)^2\left|\frac{du}{dt}\right|

Evaluating at each distance

With f=1 mf=1\ \text{m} and u=−du=-d, u−f=−(d+1)u-f=-(d+1), so (fu−f)2=1(d+1)2\left(\frac{f}{u-f}\right)^2 = \frac{1}{(d+1)^2}, and the image speed is 5(d+1)2 m s−1\frac{5}{(d+1)^2}\ \text{m s}^{-1}.

Distance ddv=fuu−fv=\frac{fu}{u-f}Image speed
39 m39\ \text{m}0.975 m0.975\ \text{m}51600=1320≈3.1×10−3 m s−1\frac{5}{1600}=\frac{1}{320}\approx3.1\times10^{-3}\ \text{m s}^{-1}
29 m29\ \text{m}0.967 m0.967\ \text{m}5900=1180≈5.6×10−3 m s−1\frac{5}{900}=\frac{1}{180}\approx5.6\times10^{-3}\ \text{m s}^{-1}
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