Physics

# The position of x of a particle at time t is given by $x = \frac { V _ { 0 } } { a } \left( 1 - e ^ { - a t } \right)$ where $V _ { 0 }$ is constant and a>0  The dimensions of $V _ { 0 }$ and a are

$M ^ { 0 } L T ^ { - 1 } \text { and } T ^ { - 1 }$

##### SOLUTION
The position of the particle at a time t is given as

$X(t)=\dfrac { { { V_{ 0 } } } }{ { a(1-{ e^{ -at } }) } }$
As X denotes position
$\left[ X \right] =m$                 here, $\left[ a \right] ={ s^{ -1 } } \rightarrow a=\left[ { { M^{ 0 } }{ L^{ 0 } }{ T^{ -1 } } } \right]$

$\Rightarrow \dfrac { { \left[ { { V_{ 0 } } } \right] } }{ { \left[ a \right] } } =\left[ { { V_{ 0 } } } \right] s=m$

Hence, ${ V_{ 0 } }$ has dimensions of velocity
$\left[ { { V_{ 0 } } } \right] =m{ s^{ -1 } } \\ { V_{ 0 } }=\left[ { { M^{ 0 } }{ L^{ 1 } }{ T^{ -1 } } } \right]$
so the correct option is $A.$

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Single Correct Medium Published on 18th 08, 2020
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