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


ANSWER

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


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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