I used my value from part ii for velocity and multiplied it with the time i got in part ii as well.I haven't had a go at the problem but from a 2 second glance, try using distance=speedxtime. Let me know if that works. If it doesn't, I'll try and do the whole problem for you
Thank you so much! That completely slipped my mind.v=0.12t(6-t)=dx/dt
Integrating with respect to time:
=>x=0.12[3t^2-0.5t^2]+C
When t=0, x=0 (note x describes the displacement (or in this case, distance) from the first station after a time t):
C=0
Hence, x=0.12[3t^2-(1/3)t^3]
From part (i) we know that when t=6 minutes the train reaches the second station so when t=6:
=>x=0.12[3*6^2-(1/3)*6^3] km
Thanks Delsa, but you don't have to do it.Allright, I'll have a go after dinner