Question

A car is fitted with a convex side-view mirror of a focal length of $20cm$. A second car $2.8m$ behind the first car is overtaking the first car at a relative speed of $15m/s$. The speed of the image of the second car as seen in the mirror of the first one is:
A) $\dfrac{1}{{10}}m/s$
B) $\dfrac{1}{{15}}m/s$
C) $10m/s$
D) $15m/s$

Answer 1

Hint: For such type of questions, you can first find the distance of the image from the mirror, then differentiate the mirror equation $\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$ to find the relation between the velocities of the image and the object and put in the values of $v,u$ and $f$ in that relation to find velocity or speed of the object.

Complete step by step solution:
We’ll follow the procedure mentioned in the hint, this is the simplest and conceptual solution, which can help in strengthening out concepts.
We know that the focal length of a convex mirror is positive. And since the object is real, we can see that the value of the distance of the object ( $u$ ) is negative.
Hence, given in the question:
$f = + 20cm$ (Focal length of the mirror)
$u = - 2.8m = - 280cm$ (Distance of the object from mirror)
${V_o} = \dfrac{{du}}{{dt}} = 15m/s$ ( ${V_o}$ = Velocity of object)
We have to find ${V_i}$ (Velocity of image)
The mirror equation using which we’ll solve the question:
$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$
First, we need to find the value of distance of image from the mirror ( $v$ ) using mirror equation:
$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$
Substituting the values that are given to us in the question:
$\dfrac{1}{{20}} = \dfrac{1}{{ - 280}} + \dfrac{1}{v}$
After transposing, we get:
$\dfrac{1}{{20}} + \dfrac{1}{{280}} = \dfrac{1}{v}$

After taking LCM and reciprocating further, we get the value of $v$ :
$\dfrac{{14 + 1}}{{280}} = \dfrac{1}{v}$
$\dfrac{1}{v} = \dfrac{{15}}{{280}}$
$v = \dfrac{{280}}{{15}}cm$
Now we need to differentiate the mirror equation with respect to time to develop the relation between velocity of image and velocity of object:
$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$
Differentiating with respect to time:
$\dfrac{{d\dfrac{1}{f}}}{{dt}} = \dfrac{{d\dfrac{1}{u}}}{{dt}} + \dfrac{{d\dfrac{1}{v}}}{{dt}}$
Since focal length of a mirror is always constant: $\dfrac{{d\dfrac{1}{f}}}{{dt}} = 0$
Using the basic differential rules:
$0 = \dfrac{{ - 1}}{{{u^2}}}\dfrac{{du}}{{dt}} + \dfrac{{ - 1}}{{{v^2}}}\dfrac{{dv}}{{dt}}$
After transposing, we get:
$\dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}} = - \dfrac{1}{{{u^2}}}\dfrac{{du}}{{dt}}$
$\dfrac{{dv}}{{dt}} = - \dfrac{{{v^2}}}{{{u^2}}}\dfrac{{du}}{{dt}}$
Putting in the value for $v$ , $u$ and $\dfrac{{du}}{{dt}}$ , we get:
$\dfrac{{dv}}{{dt}} = - \dfrac{{{{\left( {\dfrac{{280}}{{15}}} \right)}^2}}}{{{{\left( {280} \right)}^2}}} \times \left( {15} \right)$
Solving the equation, we get:
$\dfrac{{dv}}{{dt}} = - \dfrac{1}{{15}}m/s$
$Speed = \left| {\dfrac{{dv}}{{dt}}} \right| = \dfrac{1}{{15}}m/s$ (Since speed can never be negative, while velocity can be)

Hence, Option (B) is correct.

Note:
Many students take the focal length of a convex mirror as negative, which is wrong, focal length of a convex mirror is always positive. Also, since this is a very common question, so, there is a generalized formula to find the answer faster:
${V_i} = \,\, - \dfrac{{{f^2}}}{{{{\left( {u - f} \right)}^2}}}\dfrac{{du}}{{dt}}$ (You have to consider the signs of $u$ and $f$ in this equation)