The velocity v of a particle at time t is given by b v = at + b/(t+c) , where a, b and c are constants. The dimensions of a, b andc are

The correct answer is Option [LT⁻²], [L] and [T].

To find the dimensions of the constants $a$, $b$, and $c$ in the given equation, we use the principle of homogeneity of dimensions. According to this principle, the dimensions of each term on both sides of a physical equation must be the same. Only physical quantities with the same dimensions can be added or subtracted.

The given equation is:
$v=at+\frac{b}{t+c}$
where:
- $v$ is the velocity, which has the dimensions of $\left[L{T}^{-1}\right]$.
- $t$ is the time, which has the dimensions of $\left[T\right]$.

Step 1: Finding the dimensions of $c$
In the denominator of the second term, we have the expression $t+c$. Since time $t$ is added to the constant $c$, both must have the same dimensions.
Therefore, the dimensions of $c$ must be the same as the dimensions of $t$:
$\left[c\right]=\left[t\right]=\left[T\right]$

Step 2: Finding the dimensions of $a$
According to the principle of homogeneity, the dimensions of the term $at$ must be equal to the dimensions of velocity $v$:
$\left[at\right]=\left[v\right]$
$\left[a\right]\left[t\right]=\left[v\right]$
$\left[a\right]\left[T\right]=\left[L{T}^{-1}\right]$
Solving for $\left[a\right]$:
$\left[a\right]=\frac{\left[L{T}^{-1}\right]}{\left[T\right]}=\left[L{T}^{-2}\right]$

Step 3: Finding the dimensions of $b$
Similarly, the dimensions of the term $\frac{b}{t+c}$ must also be equal to the dimensions of velocity $v$:
$\left[\frac{b}{t+c}\right]=\left[v\right]$
Since the denominator $t+c$ has the dimensions of time $\left[T\right]$, we write:
$\frac{\left[b\right]}{\left[T\right]}=\left[L{T}^{-1}\right]$
Solving for $\left[b\right]$:
$\left[b\right]=\left[L{T}^{-1}\right]\cdot \left[T\right]=\left[L\right]$

Conclusion:
The dimensions of $a$, $b$, and $c$ are $\left[L{T}^{-2}\right]$, $\left[L\right]$, and $\left[T\right]$ respectively.