Question

A car is fitted with a convex side view mirror of f=20cm. A 2 nd car 2.8m behind the 1 st car is overtaking the 1 st car with a relative speed of 15m/s what will be the speed of 2 nd car w.r.t. the first one?

Answer 1

Here, consider that the first car as the object and the second car as the image. Then, using the mirror formula, find the image distance, which is the distance of the second car. Then, differentiate the mirror formula which will give the relationship between the velocities of the two cars and substitute the values.

Complete step by step solution: Here, we know that f=+20 which is the focal length. The first car is 2.8m behind, which is our object distance denoted by u, where u=-280cm. The relative speed of the first car is 15m/s, thus $\dfrac{du}{dt}=15\dfrac{m}{s}$.

Now, we use the mirror formula to find the image distance, that is, the distance of the second car is denoted by v, and v can be obtained as follows:

$$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f} \\\ \Rightarrow\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u} \\\ \Rightarrow\dfrac{1}{v}=\dfrac{u-f}{uf} \\\ \Rightarrow\dfrac{1}{v}=\dfrac{280+20}{20\centerdot (280)} \\\ \Rightarrow v=\dfrac{280}{15}cm \\\$$

Here, u is the distance of the first car, and f is the focal length of the second car’s side view mirror.

Now, the velocity of the second car will be obtained by differentiating the mirror formula with respect to time as follows:
$\dfrac{d}{dt}(\dfrac{1}{u})+\dfrac{d}{dt}(\dfrac{1}{v})=\dfrac{d}{dt}(\dfrac{1}{f})$
But the focal length of the mirror remains constant; hence its derivative will be zero.
${\dfrac{du}{dt}}(\dfrac{ - 1}{u^2}) + {\dfrac{dv}{dt}}(\dfrac{ - 1}{v^2}) = 0$
${\dfrac{ - 1}{v^2}} {\dfrac{{dv}}{{dt}}} = {\dfrac{{du}}{{dt}}}{(\dfrac{1}{{{u^2}}})}$
${\dfrac{{dv}}{{dt}}} = {\dfrac{{ - {v^2}}}{{{u^2}}}} {\dfrac{{du}}{{dt}}}$
Here,.$\dfrac{du}{dt}$. is the velocity of the first car which is 15m/s and$\dfrac{dv}{dt}$is the velocity of the second car.
Thus,

$$\dfrac{{dv}}{{dt}} = \dfrac{{ - {{(\dfrac{{280}}{{15}})}^2}}}{{{{280}^2}}} \times 15 \\\ \therefore\dfrac{{dv}}{{dt}} = \dfrac{{ - 1}}{{15}}\dfrac{m}{s}$$

Thus, the velocity of the second car is$- \dfrac{1}{{15}}\dfrac{m}{s}$.

Notes: The equation $\dfrac{dv}{dt}=\dfrac{-{{v}^{2}}}{{{u}^{2}}}\dfrac{du}{dt}$is a general equation and holds true for all types of mirrors, as the focal length of the mirror will remain constant for all the mirrors and hence its differentiation with respect to time will always be zero. This equation can also be derived for lenses and it will be of the same type, just the difference will be that there will be no negative sign.