Question

The motors of an electric train can give it an acceleration of $1{\text{m}} \cdot {{\text{s}}^{ - 2}}$ and the breaks can give a negative acceleration of $3{\text{m}} \cdot {{\text{s}}^{{\text{ - 2}}}}$ . The shortest time in which the train can make a trip between two station $1350{\text{ m}}$ apart is
A) $113.6\;{\text{s}}$
B) $60\;{\text{s}}$
C) $245.4\;{\text{s}}$
D) $14.2\;{\text{s}}$

Answer 1

Hint:
Kinematical equations of motion can provide the relationship between the different parameters of the moving object. There are three kinematical equations of motion. Use these equations to get the answer.

Step by step solution:

The first kinematical equation is given as shown below.
$ \Rightarrow v = u + at$

The second kinematical equation is given as shown below.

$ \Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
The third kinematical equation is given as shown below.
According to the question it is given that the value of an acceleration is ${a_1} = 1{\text{ m}} \cdot {{\text{s}}^{ - 2}}$, the value of deceleration is ${a_2} = - 3{\text{ m}} \cdot {{\text{s}}^{ - 2}}$ and the value of the distance is $s = 1350{\text{ m}}$.

The train starts from rest with acceleration for time ${t_1}$ for distance ${s_1}$. In the next distance ${s_2}$, the train accelerates for time ${t_2}$.
Apply kinematical equation of motion for distance ${s_1}$ and it is written as,
\[
   \Rightarrow v = u + {a_1}{t_1} \\
   \Rightarrow {s_1} = ut + \dfrac{1}{2}{a_1}t_1^2 \\
 \]
As the train is starting from rest, the initial velocity of the train is $0{\text{ m/s}}$.
So, kinematical equations are given as:
The final velocity of the train is calculated as,

\[
   \Rightarrow v = 0 + {a_1}{t_1} \\
   \Rightarrow v = {t_1} \\
 \] … (i)
The distance travelled in time ${t_1}$:
\[
   \Rightarrow {s_1} = \left( 0 \right)t + \dfrac{1}{2}{a_1}t_1^2 \\
   \Rightarrow {s_1} = \dfrac{1}{2}t_1^2 \\
 \]
For deceleration:
Kinematical equation becomes:
For distance ${s_2}$, The final velocity of the train is zero and its initial value is v.
\[
   \Rightarrow 0 = u + {a_2}{t_2} \\
   \Rightarrow {t_1} = 3{t_2} \\
 \] … (ii)
\[
   \Rightarrow {s_2} = u{t_2} + \dfrac{1}{2}{a_2}t_2^2 \\
   \Rightarrow {s_2} = u{t_2} - \dfrac{1}{2}3t_2^2 \\
 \]

Add equation of distances:
\[ \Rightarrow {s_1} + {s_2} = u{t_2} + \dfrac{1}{2}t_1^2 - \dfrac{1}{2}3t_2^2\]
Substituting the equation (ii) and (i) and it is written as,
\[
   \Rightarrow {s_1} + {s_2} = u{t_2} + \dfrac{1}{2}{\left( {3{t_2}} \right)^2} - \dfrac{1}{2}3t_2^2 \\
   \Rightarrow {s_1} + {s_2} = {t_1}{t_2} + \dfrac{9}{2}t_2^2 - \dfrac{3}{2}t_2^2 \\
   \Rightarrow {s_1} + {s_2} = \left( {3{t_2}} \right){t_2} + \dfrac{6}{2}t_2^2 \\
   \Rightarrow 1350 = 3t_2^2 + 3t_2^2 \\
   \Rightarrow 1350 = 6t_2^2 \\
   \Rightarrow {t_2} = 15\;{\text{s}} \\
 \]
Therefore,
$
   \Rightarrow {t_1} = 3{t_2} \\
   \Rightarrow {t_1} = 45\;{\text{s}} \\
 $

The shortest time is given as:
$
   \Rightarrow t = {t_1} + {t_2} \\
   \Rightarrow t = 15 + 45 \\
   \Rightarrow t = 60\;{\text{s}} \\
 $

So, the shortest time is $60\;{\text{s}}$.

Hence, the option B is correct.

Note:The deceleration is the negative acceleration. When the break on the electric train is applied, the velocity of the train decreases. This results in deceleration of the train. The term deceleration of the object causes the object to stop.