Question

A train stops at two stations P and Q which are $2\,km$ apart. It accelerates uniformly from P at $1\,m{s^2}$ for $15$ seconds and maintains a constant speed for a time before decelerating uniformly to rest at Q. If the deceleration is $0.5\,m{s^2}$ , find the time for which the train is travelling at a constant speed.
A. $29.16\,\sec $
B. $111\,\sec $
C. $121\,\sec $
D. $131\,\sec $

Answer 1

Hint: Here we have to use the first and third equation of motion to get the answer.
In physics, equations of motion are classified as equations that characterise a physical system’s behaviour in terms of its motion as a function of time. For deriving components such as displacement, velocity, time and acceleration, there are three motion equations that can be used.

Complete step by step answer:
The first equation of motion deals with final velocity, initial velocity, acceleration and time.
The first equation of motion is given by:
$v = u + at$
For first $15\,\sec $ ,
$u = 0$
$a = 1 + 1 = 2m{s^2}$
So, applying the first equation of motion we get:
$
  v = u + at \\
\implies v = u + at \\
\implies v = 0 + 2 \times 15 \\
\implies v = 30\,m{s^{ - 1}} \\
$
When the train is decelerating
$v' = 0$
Again applying the first equation of motion we get:
$
v' = u' + a't' \\
\Rightarrow 0 = 30 - \dfrac{1}
{2}t' \\
   \Rightarrow t' = 60\,\sec \\
$
Now we have to apply the third equation of motion.
The third equation of motion gives the relation between initial velocity, final velocity and distance.
It is mathematically given by:
${v^2} - {u^2} = 2as$
When the initial velocity, $u = 0$ , we get:
$ {v^2} - {u^2} = 2a{s_1} \\
\implies {30^2} - 0 = 2 \times 2 \times {s_1} \\
\implies {s_1} = 225\,m \\
$
Again applying the third equation of motion, when $v = 0$ , we get:
$
  0 - {30^2} = 2 \times \left( {\dfrac{{ - 1}}
{2}} \right) \times {s_3} \\
   \Rightarrow {s_3} = 900\,m \\
$
So, when speed is constant we get:
$
  {s_2} = 2000 - {s_3} - {s_2} \\
   \Rightarrow {s_2} = 2000 - 900 - 225 \\
   \Rightarrow {s_2} = 875\,m \\
$
So, the time when the speed is constant is:
$
  t'' = \dfrac{{{s_2}}}
{v} \\
   = \dfrac{{875}}
{{30}} \\
   = 29.16\,\sec \\
$

So, the correct answer is “Option A”.

Additional Information:
Acceleration is known to be used as a term for speeding up and deceleration for speeding down. However, speeding up may be referred to as positive acceleration and slowing down as negative acceleration.

Note:
Here we have to observe the question carefully and see when will the initial velocity be zero and when will the final velocity be zero.
We have to use both the cases in the first and third equation of motion as well.