Question

Two trains travelling on the same track are approaching each other with equal speed of $40\,m\,{s^{ - 1}}$ . The drivers of the trains begin to decelerate simultaneously when just $2.0\,km$ apart. Assuming the decelerations to be uniform and equal, the value to the deceleration to barely avoid collision should be
A. \[11.8\,m\,{s^{ - 2}}\]
B. $11.0\,m\,{s^{ - 2}}$
C. $2.1\,m\,{s^{ - 2}}$
D. $0.8\,m\,{s^{ - 2}}$

Answer 1

Hint: The above question is of relative motion in a non-uniform manner.Using equations of non-uniform straight line relative motion alone will be the best approach.A basic knowledge of relative motion and using the statement's approach one can lead to the answer. Imagining the situation will help a better understanding and easy approach.

Complete step by step answer:
Non-uniform motion in a straight line is motion of an object in a straight line with velocity of object varying with every passing second i.e. in simple words non-uniform motion is an accelerated motion. According to the question both the trains are under non-uniform motion in a straight line.

If the acceleration in the body reduces the velocity of the body it is known as deceleration. While decelerating there is reduction in velocity of body because acceleration in body is in direction opposite to the motion in body, hence in these cases acceleration is considered negative.

For solving non-uniform motion in straight line we have 3 basic equations:
$v = u + at\,\,\,\,\,.....\left( i \right)$
$\Rightarrow {v^2} - {u^2} = 2aS\,\,\,\,\,.......\left( {ii} \right)$
$\Rightarrow S = ut + \dfrac{1}{2}a{t^2}\,\,\,\,\,\,\,\,.......\left( {iii} \right)$
Where, $v = {\text{final velocity of body}}$ ; $u = {\text{initial velocity of body}}$ ; $t = {\text{time}}$; $S = {\text{displacement}}$ ; $a = {\text{acceleration in body}}$

Now if we imagine the situation we will observe that both trains are approaching each other with equal velocity of $40\,m\,{s^{ - 1}}$ . As both trains are moving in opposite directions relative to each other so their velocity vectors will add up. So the initial velocity will be given as,
$u = 40 + 40 = 80\,m\,{s^{ - 1}}$
As finally both the trains are to be halted, hence
$v = 0\,m\,{s^{ - 1}}$
Relative distance to be covered by both trains,
$S = 2.0\,km = 2000\,m$

As we need to calculate the acceleration needed to stop both trains, the appropriate equation that beholds all the given values and to final values is equation (ii),
Let relative acceleration in trains be: $(a)\,m\,{s^{ - 2}}$
Putting values in equation (ii),
${\left( 0 \right)^2} - {\left( {80} \right)^2} = 2 \times 2000 \times a$
$a = ( - 1.6)\,m\,{s^{ - 2}}$

According to question acceleration in both the trains is equal and is in opposite direction hence the net acceleration is addition of both acceleration. Let acceleration in trains be: $(a')\,m\,{s^{ - 2}}$
$a = a' + a' = 2a'$
Putting value of $a$ in above equation,
$( - 1.6) = 2a'$
$\therefore a' = ( - 0.8)\,m\,{s^{ - 2}}$
Negative sign denotes that acceleration is acting in the opposite direction, providing resistance to flow. This acceleration is known as deceleration. Deceleration produced in trains to stop both the trains is $0.8\,m\,{s^{ - 2}}$.

Hence, the correct answer is option D.

Note: It is important to note the units of values. Comparing different dimensional units will yield a negative result. The negative sign in vector values also denotes the direction of application of that vector. It doesn’t imply that the output answer is below zero value. Relative motion is very important, especially while applying the signs of the body. Always remember that opposite directions will add up and similar directions will subtract each other.