Linear motion physics question (with speed, distance)?

Question

A train moves at a constant speed of 60 km/h toward a station 30 km away. At that moment Fanny Fastbird leaves her perch on the locomotive and flies toward the station at a constant speed of 100 km/h relative to the ground. When the bird reaches the station, she immediately turns around and flies back to the train at the same speed. When reaching the train she again immediately turns around and flies back to the station, repeating the process until the train passes the station. What total distance is traveled by the bird?


Thank you so much for your help! I am curious as to how I would solve this problem, and the more steps shown, the better. Thank you!

EDIT #1: Rick, I've tried that.

The train takes 30 minutes or .5 hours to go the 30 km. The bird takes 18 minutes or .3 hours to go the 30 km.

It gets real difficult after that because the bird reverses direction, and then repeats this process at least once.

EDIT #2. if my math is right, I forgot to say, in the time it takes the bird to fly the 30 km, the train has traveled 18 km.

Answer 1

This problem can be simply stated as:

How far the bird travels with speed 100km/hr in the time taken by the train to reach the station 30 km away with speed 60km/hr

So the time taken by the train = 30/60 = 0.5 hour

Distance travelled by the bird during this time will be

100*0.5 = 50km

Answer 2

This is a classic question, and the solution is easier than it looks.

First, figure out how long it takes the train to reach the station (hint: t = distance/speed).

Next, figure out how far a bird, traveling at 100 km/h, can travel during that amount of time.

Answer 3

ok, there's a trick to this utilising reality it somewhat is now no longer your standard projectile action question. in case you have been in basic terms knocking a flower pot off the roof, then you definately easily might choose to persist with standard equations of their least perplexing type. yet, you're throwing the brick UP at an perspective. So, first you will possibly desire to persist with trig to locate the preliminary vertical speed: sine(25 deg) = X / 15 m/s, the area X is the vertical speed. then you definately easily use V(very very final) = V(preliminary) - gt, the area g is gravitational acceleration, t is time, and V(very very final) is 0. this might grant you the quantity of time the brick took to attain the utmost good in its upward arc. then you definately easily use: y = vt - (a million/2)gt^2, the area y is vertical distance, v is preliminary speed (the vertical ingredient you calculated), g is gravitational acceleration, and t is the time it took to attain the authentic of its arc (which you already calculated). this might grant you the vertical distance that the brick traveled upward until eventually now all of it began out its downward plunge. then you definately easily take it sluggish you calculated for the upward action and subtract it from the three seconds. This answer will grant you the quantity of time the brick traveled downward. Now. in case you make the main of the equation from above ( y = vt - (a million/2)gt^2 ) the area y sticks out because of the fact the vertical fall distance, vt is 0 utilising reality the preliminary vertical speed of the brick from the best of its arc is 0, and t is the time which you get as quickly as you subtract your calculated upward time from the three seconds, then you definately easily gets the carried out vertical FALL distance. utilising reality the hollow the brick fell grew to become from an excellent extra effective than the best of the form (it grew to become thrown upward), you will might desire to subtract the "upward" vertical distance which you until eventually finally now calculated from this new distance as a thank you to get the properly suited good of the form.