Question

A train starts from rest and moves with a constant acceleration of $2.0m{s^{ - 2}}$ for half a minute. The brakes are then applied and the train comes to rest in one minute. Find
(a) the total distance moved by the train,
(b) the maximum speed attained by the train and
(c) the position(s) of the train at half the maximum speed.

Answer 1

Using the first half minute we can find the final velocity of the train and after that the train applies brakes and comes to rest hence this is the maximum speed of the train and the distance traveled by the train. After that we can find the deceleration of the train and from that we can calculate the distance traveled by deceleration now on adding both the distance to find the total distance. Now during the same calculation with half the maximum speed we can find the position of the train.

Complete step by step answer:
As per the problem we know a train starts from rest and moves with a constant acceleration of $2.0m{s^{ - 2}}$ for half a minute. The brakes are then applied and the train comes to rest in one minute.
Now here we know,
$u = 0m{s^{ - 1}}$
$a = 2m{s^{ - 2}}$
$t = \dfrac{1}{2}\min = 30\sec$
Now using rectilinear motion,
$v = u + at$
Putting the know values we will get,
$v = 0 + 2m{s^{ - 2}}\left( {30s} \right)$
$\Rightarrow v = 60m{s^{ - 1}}$
As the train applied brakes and came to rest in one minute then there must be some deceleration of the train.
Here we know,
$u = 60m{s^{ - 1}}$
$v = 0m{s^{ - 1}}$
$t = 1\min = 60\sec$
Now using rectilinear motion,
$v = u + at$
Putting the know values we will get,
$0m{s^{ - 1}} = 60m{s^{ - 1}} + a\left( {60s} \right)$
$\Rightarrow - 60m{s^{ - 1}} = 60s \times a$
$\Rightarrow a = - 1m{s^{ - 2}}$
(a) Now we know the acceleration and deceleration acting on the body now using the distance formula of rectilinear motion we can find the distance travelled by the train.
Case I:
When the train is accelerating with $2.0m{s^{ - 2}}$.
Using rectilinear formula we will get,
$s = ut + \dfrac{1}{2}a{t^2}$
Putting the known values we will get,
$s = \left( 0 \right)t + \dfrac{1}{2}\left( {2m{s^{ - 2}}} \right){\left( {30\operatorname{s} } \right)^2}$
$\Rightarrow s = 900m$
Case II:
When the train decelerates with $1.0m{s^{ - 2}}$.
Using rectilinear formula we will get,
$s = \left( {60m{s^{ - 1}}} \right)\left( {60\operatorname{s} } \right) + \dfrac{1}{2}\left( { - 1m{s^{ - 2}}} \right){\left( {60s} \right)^2}$
Putting the known values we will get,
$s = 3600m - 1800m$
$\Rightarrow s = 1800m$
Hence the total distance is equals by adding two cases we will get,
$S = 900m + 1800m = 2700m$
(b) From equation $\left( 1 \right)$ we can say that the maximum speed attended by the train is $60m{s^{ - 1}}$.
(c) Now taking the half the maximum speed we can find the distance as,
When,
$a = 2m{s^{ - 2}}$
$t = 30s$
$u = 0m{s^{ - 1}}$
And $v = 30m{s^{ - 1}}$
Using rectilinear formula we will get,
${v^2} - {u^2} = 2as$
Putting the values we will get,
${\left( {30m{s^{ - 1}}} \right)^2} - {\left( {0m{s^{ - 1}}} \right)^2} = 2\left( {2m{s^{ - 2}}} \right)s$
$\Rightarrow s = 225m$

Note: Here we are using the formula of rectilinear motion. Remember there are three formula of rectilinear motion
(I)$s = ut + \dfrac{1}{2}a{t^2}$
(II)$v = u + at$
(III)${v^2} = {u^2} + 2as$
Rectilinear motion is defined as when a particle is restricted to move along a straight line or can be described using one coordinate only.