When dealing with problems involving speed, time, and distance, it’s important to set up equations based on the formula \( \text{Distance} = \text{Speed} \times \text{Time} \). For problems involving upstream and downstream travel, remember that the effective speed is altered by the speed of the stream. In this case, the upstream speed is reduced and the downstream speed is increased due to the current. Simplify the equation step by step to find the unknown distance, ensuring the units of time and speed match up correctly.
Given:
Speed in still water \(v = 5 \text{ km/hr}\),
Speed of stream \(u = 2 \text{ km/hr}\),
Additional time taken upstream = 20 minutes = \(\frac{1}{3}\) hours.
Effective speeds:
Speed Upstream = \(v - u = 5 - 2 = 3 \text{ km/hr}\),
Speed Downstream = \(v + u = 5 + 2 = 7 \text{ km/hr}\).
Let the distance between the two points be \(d \text{ km}\). Time taken for upstream and downstream travel:
Time Upstream = \(\frac{d}{3}\),
Time Downstream = \(\frac{d}{7}\).
The time difference between upstream and downstream travel is:
\(\frac{d}{3} - \frac{d}{7} = \frac{1}{3}\).
Equation simplification:
\(\frac{7d - 3d}{21} = \frac{1}{3}\),
\(\frac{4d}{21} = \frac{1}{3}\).
Multiplying by 21:
\(4d = 7\), \(d = \frac{7}{4} = 1.75 \text{ km}\).
Therefore, the distance between the two points is 1.75 km.
A flight, traveling to a destination 11,200 kms away, was supposed to take off at 6:30 AM. Due to bad weather, the departure of the flight got delayed by three hours. The pilot increased the average speed of the airplane by 100 km/hr from the initially planned average speed, to reduce the overall delay to one hour.
Had the pilot increased the average speed by 350 km/hr from the initially planned average speed, when would have the flight reached its destination?