Question 805738: Initially stationary (the train starts at rest), a train has a constant acceleration of 0.3m/s^2.
a)What is its speed after 34s?
b)What would be the total time it would take to reach a speed of 31m/s?
Answer by trumpettuba(3)   (Show Source):
You can put this solution on YOUR website! You can use uvast equations:
s = distance (metres, m)
u = initial velocity (metres per second, ms^-1)
v = final velocity (metres per second, ms^-1)
a = acceleration (metres per second squared, ms^-2)
t = time (seconds, s)
v = u + at
s = ut + (1/2)(a)(t^2)
s = vt - (1/2)(a)(t^2)
v^2 = u^2 + 2as
s = ((u + v)/2)t
We can use v = u + at : v=?, u=0 (since it was at rest),a=0.3m/s^2,t=34s
Sub that into the formulae v=0+(0.3)(34)Thats the answer to part (i)
For part (ii) you use the same formulae but don't sub in time as that's what we're looking for but sub in v i.e. 31m/s. That will give the answer for part (ii)
This question is answered in physics style.
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