A car driver driving in a fog passes a pedestrian who was walking at the rate of $2km/hr$ in the same direction. The pedestrian could see the car for 6 minutes and it was visible to him upto a distance of 0.6 km. What was the speed of the car?
A. $15km/hr$
B. $30km/hr$
C. $20km/hr$
D. $8km/hr$

Answer Verified Verified
Hint: Assume a variable for the speed of the car (say V), and calculate the distance covered by it in 6 minutes.
Then calculate the distance covered by the pedestrian in 6 min. As pedestrians can see up to 0.6km, the difference between the distance covered by the car and the pedestrian will be 0.6 km. Equate the difference to 0.6 km to get the equation in ‘V’ and then solve for ‘V’.

Complete step-by-step answer:
According to the question, the speed of pedestrians (in the same direction as cars) is $2km/hr$.
We have to find the speed of the car. So, let us assume the speed of the car to be $Vkm/hr$.
In 6 minutes,
Distance covered by pedestrian $=2km/hr\times \dfrac{6}{60}hr$.
And distance covered by car $=Vkm/hr\times \dfrac{6}{60}hr$.
  & 60\min =1\text{hour} \\
 & \text{1min =}\dfrac{1}{60}\text{hour} \\
 & \Rightarrow \text{6min = }\dfrac{6}{60}\text{hour} \\
Now, according to the question, pedestrians can see up to 0.6 km and 6min.
Let the car passed pedestrian at A,

By the figure, we can write that,
  & \left( \text{distance covered by car} \right)-\left( \text{distance covered by pedestrian} \right)=0.6km \\
 & \Rightarrow \left( V\times \dfrac{6}{60} \right)-\left( 2\times \dfrac{6}{60} \right)=0.6 \\
 & \Rightarrow \dfrac{V}{10}-\dfrac{2}{10}=\dfrac{6}{10} \\
Taking constant term RHS, we will get,
  & \Rightarrow \dfrac{V}{10}=\dfrac{6}{10}+\dfrac{2}{10} \\
 & \Rightarrow \dfrac{V}{10}=\dfrac{8}{10} \\
Multiplying both sides by 10, we will get,
$\Rightarrow V=8$
Hence, the required speed of the car will be $8km/hr$ and option (D) is the answer.

Note: The pedestrian can see up to 0.6 km and upto 6 min. Don’t forget to subtract the distance covered by pedestrians from the distance covered by car because pedestrians are also moving in the same direction and approaching the car.