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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
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