Given \( \frac{d v}{d t}=\frac{4 t^{2}+4 t-15}{2 t-3} \), and \( v=5 \mathrm{~ms}^{-1} \) when \( t=1 \mathrm{~s} \) determine \( v \) as a function of \( t \)

To determine \( v \) as a function of \( t \), we start by integrating the given derivative \( \frac{d v}{d t}=\frac{4 t^{2}+4 t-15}{2 t-3} \). 1. **Simplify the expression**: First, let's perform polynomial long division on the rational function: \[ \frac{4t^2 + 4t - 15}{2t - 3} \] Dividing \( 4t^2 \) by \( 2t \) gives \( 2t \). Multiply \( 2t \) by \( 2t - 3 \) to get \( 4t^2 - 6t \). Subtract this from \( 4t^2 + 4t - 15 \): \[ (4t^2 + 4t - 15) - (4t^2 - 6t) = 10t - 15 \] Now, divide \( 10t - 15 \) by \( 2t - 3 \): The result is \( 5 \) with a remainder of \( 0 \): \[ 10t - 15 - (5)(2t - 3) = 0 \] Therefore, \[ \frac{4t^2 + 4t - 15}{2t - 3} = 2t + 5 \] 2. **Integrate**: \[ v = \int (2t + 5) \, dt = t^2 + 5t + C \] where \( C \) is a constant. 3. **Determine Constant \( C \)**: We know \( v = 5 \) when \( t = 1 \): \[ 5 = (1)^2 + 5(1) + C \] \[ 5 = 1 + 5 + C \] \[ C = 5 - 6 = -1 \] 4. **Final function**: Putting it all together, the function for \( v \) is: \[ v(t) = t^2 + 5t - 1 \] To summarize, the velocity as a function of time \( t \) is given by: \[ v(t) = t^2 + 5t - 1 \]