In a journey of three unequal laps, a car covers a distance of 200 km in 4 h in the first lap, while another 162 km at the speed of 15 m/s in the second lap. It covered the remaining distance of the final lap in 4 h such that the average speed of the car for entire journey was 50 km/h. What was the speed of the car in the third lap of the journey?
Answer & explanation
Correct answer: option 1
Distance covered in first lap = 200 km, Time taken = 4 h
Distance covered in second lap = 162 km, speed = 15 m/s
= 15 × \(\frac{18}{5}\)
= 54 km/h
Time taken = \(\frac{162}{54}\) = 3h
Let the distance covered in the 3rd lap = S km
Time taken = 4h
Average Speed = \(\frac{200 + 162 + S}{4 + 3 + 4}\)
According to question ,
\(\frac{200 + 162 + S}{4 + 3 + 4}\) = 50km/h
362 + x = 50 × 11
x = 550 - 362 = 188 km
So, the speed in 3rd lap = \(\frac{188}{ 4}\)
= 47km/h
Therefore, the speed of the car in the third lap of the journey is 47 km/h