Question

the engineer of a passenger train travelling at 100 ft/ sec sights a freight train whose caboose is 600 ft. ahead on the same track. the freight train travelling at 30 ft/ sec on the same track as the passenger. the engineer of the passenger train immediately applies brakes, causing a constant deceleration of 4 ft/sec squared, while the other train continues with constant speed. a) will there be a collision? b) if so, where will it take place?

Answer 1

Find out how long it takes for the passenger train to stop. \[V=v_0+at\] Where \[v_0=100\] and \[a= -4\]

Answer 2

Should give you about 25 secs

Answer 3

Now plug that into \[x= v_ot + 1/2at^2\]

Answer 4

you should get the train travels 1250 m before stopping

Answer 5

Now since the passenger train is 600 m behind the freight train we can say the freight trains initial position is 0. While the passenger train's initial position is -600

Answer 6

So we get that at 25s the passanger train comes to rest at 650 m. (1250-600)

Answer 7

Now plug 650 m into \[x=v_0+1/2at^2\] for the freight train

Answer 8

so, there will be a collision?

Answer 9

Since it is moving at a constant velocity you know there is no acceleration. so the equation is simply \[x=v_0t\]

Answer 10

Which you solve for t by doing x(650)/(30)= which gives you 21.6667

Answer 11

So no there is no collision because the freight train was at 650m at 21.66667 s. While the passenger stopped there at 25 s

Answer 12

thank you very much, i am totally at loss. this is my daughter's assignment. she's grade 9

Answer 13

Haha no problem I'm glad to help.

Answer 14

Hi...Since you are helping your daughter and she is only in grade 9, you should both sit down and watch a series of youtube videos...this kind of question is called a kinematics equation and these videos will help you tremendously because he goes very slowly and repeats multiple times! hope this helps! :o)
https://www.youtube.com/watch?v=7J9neu_58Zc