Question

A boggy or uniformly moving train is suddenly detached from the train and stops after covering some distance. The distance covered by the boggy and distance covered by the train in the same time has relation
(a) Both will be equal
(b) First will be half of second
(c) First will be of the second
(d) No define ratio

Answer 1

Hint:In order to calculate the relation, we need to use the laws of motion(uniform acceleration) in this question. The formulae for the laws of motions (uniform acceleration) are,
\[v=u+at\],
\[s=\dfrac{1}{2}\left( u+v \right)t\],
\[s=ut+\dfrac{1}{2}a{{t}^{2}}\] and
\[{{v}^{2}}={{u}^{2}}+2as\].
Now, let$a$ be the retardation of boggy then distance covered by it be $S$. If $u$is the velocity of boggy after detaching from the train (i.e. uniform speed of train).
Also, ${{S}_{b}}$ be the speed of the boggy and ${{S}_{t}}$be the speed of the train.
Then, using laws of motion, we get
\[{{v}^{2}}={{u}^{2}}+2as\]
$\Rightarrow 0={{u}^{2}}-2as$
$\Rightarrow {{S}_{b}}=\dfrac{{{u}^{2}}}{2a}$
Time taken by the boggy to stop is,
\[v=u+at\]
\[\Rightarrow 0=u-at\]
\[\Rightarrow t=\dfrac{u}{a}\]
In this time t distance travelled by train is,
${{S}_{t}}=ut=\dfrac{{{u}^{2}}}{a}$
Hence, the ratio is $\dfrac{{{S}_{b}}}{{{S}_{t}}}=\dfrac{1}{2}$.

So, the correct answer for this question is option (b).

Additional Information:In case of uniform acceleration, there are three equations of motion which are also known as the laws of constant acceleration. Hence, these equations are used to derive the components like displacement(s), velocity (initial and final), time(t) and acceleration(a). Therefore, they can only be applied when acceleration is constant and motion is a straight line.

Note:While solving this question, one should have to know the laws of motion (uniform acceleration) and also its application. The formula used to solve this question is on laws of motions (uniform acceleration). All the formulas are stated above for reference.