Question

A starts from rest and moves with acceleration ${{a}_{1}}$. Two seconds later, B starts from rest and moves with an acceleration ${{a}_{2}}$. If the displacement of A in the ${{5}^{th}}$ second is the same as that of B in the same interval, the ratio of ${{a}_{1}}$ to ${{a}_{2}}$ is
     $A)\text{ }9:5$
     $B)\text{ 5}:9$
     $C)\text{ 1}:1$
     $D)\text{ 1}:3$

Answer 1

Hint: This problem can be solved by using the direct formula for the displacement covered by a body in its ${{n}^{th}}$ second of motion when it is under constant acceleration. Since B starts two seconds later, we have to equate the displacement of A in the fifth second to the displacement of B in the third second of its motion.
Formula used:
${{S}_{n}}=u+\dfrac{1}{2}a\left( 2n-1 \right)$

Complete answer:
We will use the formula for the displacement of a body undergoing a motion with constant acceleration in the ${{n}^{th}}$ second of its motion.
The displacement ${{S}_{n}}$ of a body undergoing motion with constant acceleration $a$, in the ${{n}^{th}}$ second of its motion is given by
${{S}_{n}}=u+\dfrac{1}{2}a\left( 2n-1 \right)$ --(1)
Where $u$ is the initial velocity of the body.
Now, let us analyze the question.
Since, both the bodies start from rest, their initial velocities of A and B are ${{u}_{A}}=0$ and ${{u}_{B}}=0$ respectively.
Let the displacement of A in the fifth second be ${{S}_{A}}$.
Since, B starts two seconds later, the displacement of B in that second will be nothing but its displacement in $5-2={{3}^{rd}}$ second of its motion.
Let this displacement be ${{S}_{B}}$.
The accelerations of A and B are ${{a}_{1}}$ and ${{a}_{2}}$ respectively.
Therefore, using (1), we get
${{S}_{A}}=0+\dfrac{1}{2}{{a}_{1}}\left( 2\left( 5 \right)-1 \right)=\dfrac{1}{2}{{a}_{1}}\left( 10-1 \right)=\dfrac{9}{2}{{a}_{1}}$ --(2)
Also, using (1), we get
${{S}_{B}}=0+\dfrac{1}{2}{{a}_{2}}\left( 2\left( 3 \right)-1 \right)=\dfrac{1}{2}{{a}_{2}}\left( 6-1 \right)=\dfrac{5}{2}{{a}_{2}}$ --(3)
Now, according to the question, the displacement of A in the fifth second of its motion and the displacement of B in that time is equal.
$\therefore {{S}_{A}}={{S}_{B}}$
Putting (2) and (3) in the above equation, we get
$\dfrac{9}{2}{{a}_{1}}=\dfrac{5}{2}{{a}_{2}}$
$\Rightarrow 9{{a}_{1}}=5{{a}_{2}}$
$\Rightarrow \dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{5}{9}$
$\Rightarrow {{a}_{1}}:{{a}_{2}}=5:9$
Therefore, we have got the required ratio of ${{a}_{1}}$ to ${{a}_{2}}$ as $5:9$.

Therefore, the correct answer is B.

Note:
We could have also solved this problem by subtracting the displacement in four seconds from the displacement in five seconds for A and the displacement in two seconds form the displacement in three seconds for B to get the respective displacements of A and B in the fifth and thirds seconds of their motion. This would have allowed us to do away with the direct formula for the displacement in the second of motion. However, using the formula made our calculations simpler. But nevertheless, it is still possible to solve this problem if the student does not remember the formula by applying this alternate method.