Find solutions to your problems with the help of IDNLearn.com's expert community. Discover the information you need quickly and easily with our reliable and thorough Q&A platform.
Sagot :
If a train starts from rest and accelerates uniformly until it has traveled 7.1km and acquired a forward velocity of 36m/s. the train then moves at a constant velocity of 36m/s for 2.1min. The train then slows down uniformly at 0.072m/s^2, until it is brought to a halt. Then the entire process take (in minutes ), approximately 17.1 minutes.
How to do the calculation of the above question?
Here, we have the following values,
u = 0
v = 36 m/s
s = 7.1 which is 7100 m
now we know that [tex]v^{2} = u^{2} + 2at[/tex]
As u = 0 hence the equation we have is [tex]v^{2} = 2at[/tex]
Now to find acceleration (a), a = [tex]\frac{36 *36}{2*7100}[/tex] = 0.091 m/[tex]s^{2}[/tex]
Now we know that a = 0.072 v = 0 u = 36 m/s then for t
v = u+at
[tex]\frac{36}{0.072}[/tex] = 500 sec
Putting the values in v = u +at
t' = (36)/(0.09) = 400 seconds
t'' = 2.1 min which is 126 seconds
and t''' is 500 seconds
Adding them up gives us the total time, which is
Total time = 400 +120+500
= 1026 second
=17.1 minutes
To know more about motion, visit:
https://brainly.com/question/3421105
#SPJ4
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Your search for solutions ends here at IDNLearn.com. Thank you for visiting, and come back soon for more helpful information.
