Question

A pigeon flies at 36 km/h to and fro between two cars moving toward each other on a straight road, starting from the first car when the car separation is 40 km. The first car has a speed of 16 km/h and the second one has a speed of 25 km/hr. By the time the cars meet head on, what are the (A) total distance and (B) net displacement flown by the pigeon?

Answer 1

Hint: Use the concept of relative motion (i.e. stop one car and see the motion with respect to only one car) and apply formulas \[d = s \times t\] (Where d =distance, s= speed and t = time) to solve the above problem.

Complete step-by-step answer:
Given that,
 Distance between the Cars = 40 km .
Relative Speed of the Car = Speed of the First Car + Opposite Speed of the Second Car with direction.
(For relative motion we stop one car and give the velocity to the other car by reversing its direction. Since in this case both the cars are moving in opposite direction, hence both get added. )
= 16 + 25
= 41 km/hr.
Now calculating the time required by the formula,
$ \Rightarrow {\kern 1pt} T = \dfrac{d}{t} = \dfrac{{40}}{{41}}\,\,\,hr$ (Where d =distance, s= speed and t = time)

(A) Total distance:
distance travelled by the bird in this time = speed of bird x time
\[
   \Rightarrow 36 \times \dfrac{{40}}{{41}} \\
   \Rightarrow 35.12\,\,km \\
 \]

Hence the total Distance covered by the Bird is 35.12 km.

(B) net displacement flown by the pigeon:
Distance traveled by the First Car in this time period = speed of first car x time.
\[
   \Rightarrow 16 \times \dfrac{{40}}{{41}} \\
   \Rightarrow 15.61\,\,km \\
 \]

During this duration both cars meet at this distance.

∴ Net Displacement = 15.61 km.

Note
i) While finding out relative speed, it is important to note that one car is assumed at rest and the opposite velocity of the other car is taken and added with direction.
ii) Displacement covered by the Car = Shortest distance between starting point to ending point.