Steven W. answered 10/15/16
Tutor
4.9
(4,296)
Physics Ph.D., college instructor (calc- and algebra-based)
Hi Prashant!
I am going to assume you mean the train speeds up with uniform acceleration 3 m/s2 (otherwise, it could not move from rest and achieve a max velocity), and then slows down with uniform acceleration of 2 m/s2, then coming to rest. If this is incorrect, then I would need to have you re-transcribe the problem for me. I also assume the time is 1 minute, or 60 seconds, in total for the trip.
Standard kinematic analysis assumes one uniform acceleration, so we have to look at the "speeding up" and "slowing down" parts individually. I assume that the acceleration switches instantly at the change from speeding up to slowing down, so that the final velocity of speeding up and the initial velocity of slowing down are the same velocity.
With that in mind, let's look at "speeding up" first. As always, if we want to use the kinematic equations to solve for one kinematic quantity, we need to know at least three others. So let's determine what we want to find (which is usually directly stated) and what we know in the speeding up phase:
SPEEDING UP
to find: (x-xo) (displacement)
know: vo (initial velocity = 0; starts from rest), a (= 3 m/s2; given; it is positive assuming train is moving in positive direction)
But that is only two things we know, so we cannot immediately solve for the displacement looking at just speeding up. So let's take a look at slowing down:
SLOWING DOWN
to find: (x-xo)
know: v (final velocity = 0; comes to rest), a (= -2 m/s2, since the train still moves in the positive direction)
But that is only two things we know in this phase, as well. So we are once again stuck.
However, we can appeal to two other pieces of information given or implied:
-- as mentioned above, the final velocity of the "speeding up" phase and the initial velocity of the "slowing down" phase are identical. We can call that velocity vc (for "common")
-- the total time of the trip is 60 s, so the sum of the time speeding up and time slowing down must be 60 s
So we can try stitching the two intervals together into something where we have enough equations to deal with our too-large (at present) number of unknowns.
[NOTE: from here on, I will label any quantities in the speeding up phase with subscript "u," and those in the slowing down phase with "d"]
We have two unknowns, (x-xo)u and (x-xo)d, to solve for. So we need at least two equations. But we also have three other unknowns floating around: vc, tu, and td. Let's try to reduce that number of unknowns using the conditions we mentioned above. Since we have a relationship of the final velocity of the speeding up phase to the initial velocity of the slowing down phase, and a relationship between the times tu and td, let's try:
vu = vou + autu for the speeding up phase (remember, vou = 0, since it starts at rest)
vd = vod + adtd for the slowing down phase (remember, vd = 0, since it ends at rest)
Since vu (the final speeding up velocity) and vod (the initial slowing down velocity) both equal vc, we can rewrite these as:
vc = 0 + (3 m/s2)tu = 3tu
0 = vc+(-2 m/s2)td --> vc = 2td
Since vc = 3tu and vc = 2td, we can equate the expression for the two times:
3tu = 2td
In addition, we know that tu + td = 60 s.
We can use these last two equations, with two unknowns, to solve for either time. For no particular reason, I choose tu. So:
td = 60 - tu
3tu = 2(60-tu) = 120-2tu
5tu = 120
tu = 24 s
Once we have solved for tu, we can also solve for td immediately. Then, those times become the third kinematic quantity we know for each phase, and we can solve for the displacement in each phase, then add them together to get the total displacement.
I hope that helps get you on your way! If you have any further questions about how to complete this, or would like to check an answer, just let me know.