A train is travelling at 30 mph towards a terminus which is 120 miles away.
At the same time, a pigeon starts flying from the terminus towards the train.
When it meets the train, it turns and flies back towards the terminus and, when it arrives there, it turns and flies back towards the train again. The pigeon flies at an average speed of 40 mph.
The pigeon continues to do this until the train reaches the terminus.
Assuming that no time was lost in each turnaround, how far (to the nearest 10 miles) does the pigeon fly altogether?
A. 160 miles
B. 180 miles
C. 240 miles
D. 140 miles
E. 120 miles
Determine what they are trying to ask, and what you will need to solve it
Eliminate any non-essential information
Draw a picture, graph, or equation
In moments of high stress like exam taking, always work with the paper they give you to avoid careless mistakes.
Solve.
This is a question students commonly overthink and end up wasting a lot of time on when in reality it’s very simple. We do not need to draw any complex diagrams showing the pigeon going back and forth and then calculating the distance every time…this results in a mess. Instead, look at the bigger picture. We are given the speed of the train and pigeon, the distance the train travels, and the key point that the pigeon does not slow down in a turnaround. So our approach is:
Calculate the time it takes for the train to travel to the terminus
\frac{120 miles}{30mph} = 4h
Calculate how far the bird goes during this time
(4h)(40mph) = 160 miles
Notice how simple the problem is if we calculate the problem overall instead of trying to break it into each back and forth segment the pigeon forms with the train?
\fcolorbox{red}{grey!30}{Therefore the bird travels $160$ miles in this time.}