Question

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. how long does the entire process take (in minutes )?

Answer 1

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 $v^{2} = u^{2} + 2at$

As u = 0 hence the equation we have is $v^{2} = 2at$

Now to find acceleration (a), a = $\frac{36 *36}{2*7100}$ = 0.091 m/$s^{2}$

Now we know that a = 0.072 v = 0 u = 36 m/s then for t

v = u+at

$\frac{36}{0.072}$ = 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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