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# Question #cb7bf

Mar 30, 2017

$a \to \left[\frac{L}{T \cdot T}\right]$

$b \to \left[\frac{L}{T}\right]$

$c \to \left[T\right]$

#### Explanation:

I am considering the equation as :

$v = a t + b \cdot \frac{t}{t + c}$

Let me know if this is not what you meant.

In any equation, each term separated by $+$ or $-$ must have same dimension or else cannot be added.

So here $v$ , $a \cdot t$ , and $b \cdot \frac{t}{t + c}$ must have same dimension. Since we know the dimension of velocity as $\left[\frac{L}{T}\right]$, the other 2 terms should also have the same dimension.

So $a \cdot t$ has dimension $\left[\frac{L}{T}\right]$

$\left[a T\right] = \left[\frac{L}{T}\right]$

$\left[a\right] = \left[\frac{L}{T \cdot T}\right]$

Similarly, let us consider the denominator part of 2nd term. $\left(t + c\right)$. Here $t$ and $c$ must have the same dimensions since they are being added, so

$\left[c\right] = \left[T\right]$

The rest

$\left[b \frac{T}{c + T}\right] = \left[\frac{L}{T}\right]$

$\left[b \frac{T}{T}\right] = \left[\frac{L}{T}\right]$, since $c$ and $T$ are being added

$\left[b\right] = \left[\frac{L}{T}\right]$

Hope you have a good day.

Mar 30, 2017

The dimension of $a$ and $b$ is $= \left[L\right] {\left[T\right]}^{-} 1$
The dimension of $c$ is $\left[T\right]$

#### Explanation:

The dimension of $c$ is $\left[T\right]$

The dimension of $v$ is $\left[L\right] {\left[T\right]}^{-} 1$

The dimension of $a t$ is $\left[L\right]$

The dimension of $a$ is $\left[L\right] {\left[T\right]}^{-} 1$

The dimension of $b$ is $\left[L\right] {\left[T\right]}^{-} 1$

I considered the equation as

$v = \frac{\left(a t + b t\right)}{\left(t + c\right)}$

$L H S$ is $= \left[L\right] {\left[T\right]}^{-} 1$

$R H S$ is $= \frac{\left[L\right] {\left[T\right]}^{-} 1}{T} \cdot \left[T\right]$

Therefore,

$a$ is $\left[L\right] {\left[T\right]}^{-} 1$

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