Velocity v and acceleration a in two systems of units 1 and 2 are related as v 2 = n m 2 v 1 and a 2 = a 1 m n respectively. Here m and n are constants. The relations for distance and time in two systems respectively are

Distance is stated as [ L ] = v 2 [ a ] ⇒ v 2 2 a 2 = n m 2 v 1 2 a 1 m n 2 = n 3 m 3 v 1 2 a 1 Or, we have L 2 = n 3 m 3 L 1 Similarly, time T = v a ⇒ v 2 a 2 = n m 2 v 1 a 1 m n = n 2 m v 1 a 1 So, T 2 = n 2 m T 1

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PhysicsKinematics1-DMedium2 minPYQ_2022

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Velocity $(v)$ and acceleration $(a)$ in two systems of units $1$ and $2$ are related as $v_{2} = \frac{n}{m^{2}} v_{1}$ and $a_{2} = \frac{a_{1}}{m n}$ respectively. Here $m$ and $n$ are constants. The relations for distance and time in two systems respectively are

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Solution:

Distance is stated as $[L] = \frac{[v^{2}]}{[a]}$

$\Rightarrow \frac{{[v_{2}]}^{2}}{[a_{2}]} = \frac{[{\frac{n}{m^{2}} v_{1}|}^{2}}{{[\frac{a_{1}}{m n}]}^{2}}$ $= \frac{n^{3}}{m^{3}} \frac{{[v_{1}]}^{2}}{[a_{1}]}$

Or, we have $[L_{2}] = \frac{n^{3}}{m^{3}} [L_{1}]$

Similarly, time $[T] = \frac{[v]}{[a]}$

$\Rightarrow \frac{[v_{2}]}{[a_{2}]} = \frac{[\frac{n}{m^{2}} v_{1}]}{[\frac{a_{1}}{m n}]} = \frac{n^{2}}{m} \frac{[v_{1}]}{[a_{1}]}$

So, $T_{2} = \frac{n^{2}}{m} T_{1}$

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:1-D

⏱ 2mℹ️ Source: PYQ_2022

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