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# The velocity n m/s of a particle is proportional to the cube of the time. If the velocity after 2s is 4m/s, then n is equal to:(A) ${t^3}\;$(B) $\dfrac{{{t^3}\;}}{2}$(C) $\dfrac{{{t^3}\;}}{3}$(D) $\dfrac{{{t^3}\;}}{4}$

Last updated date: 12th Sep 2024
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Hint It should be known to us that velocity is defined as the rate of change of the position of an object with respect to the frame of reference of the time. Based on this concept we have to solve this question.

The velocity of a particle is given as n m/s and it is proportional to the cube of the time.
Given that the velocity becomes 4m/s after 2 seconds,
We can derive that,
$n{\text{ }}\propto {\text{ }}{\left[ {Time\left( t \right)} \right]^3}$
Therefore, we can say that,
$n = k{t^3} \ldots \ldots \ldots \left( {\text{i}} \right)$
When velocity after 2 seconds become 4m/s,
We can equate,
$4 = k{\left( 2 \right)^3} \ldots \ldots \ldots \left( {{\text{ii}}} \right)$
$\Rightarrow 4 = 8k$
$\Rightarrow k = \dfrac{1}{2}$
Hence, we can derive n from the above equations as
${\text{n = }}\dfrac{{{{\text{t}}^{\text{3}}}}}{{\text{2}}}$ (from equation i)

Therefore, the correct answer is Option B.

Note To find the velocity we should always remember that we have to divide the change in the position by the change in time. Since it is a vector quantity we have to mention the direction as well.