1. Given relation

We have $t=ax^2+bx$ where $a$ and $b$ are constants.

We need to find the acceleration.

2. Differentiate to get velocity

Since $t$ is given as a function of $x$, differentiate with respect to $x$: $\frac{dt}{dx}=2ax+b$

But $v=\frac{dx}{dt}=\left(\frac{dt}{dx}\right)^{-1}$ So, $v=\frac{1}{2ax+b}$

3. Find acceleration

Acceleration is $a_{\text{acc}}=\frac{dv}{dt}$ Using chain rule, $\frac{dv}{dt}=\frac{dv}{dx}\cdot\frac{dx}{dt}=v\frac{dv}{dx}$

Now, $v=(2ax+b)^{-1}$ Differentiate with respect to $x$: $\frac{dv}{dx}=-1(2ax+b)^{-2}(2a)=-\frac{2a}{(2ax+b)^2}$

Thus, $a_{\text{acc}}=v\frac{dv}{dx}$ $a_{\text{acc}}=\frac{1}{2ax+b}\left(-\frac{2a}{(2ax+b)^2}\right)$ $a_{\text{acc}}=-\frac{2a}{(2ax+b)^3}$

Since $v=\frac{1}{2ax+b}$ we get $v^3=\frac{1}{(2ax+b)^3}$

Hence, $a_{\text{acc}}=-2av^3$

4. Compare with options

• A: $2bv^3$ ❌
• B: $-2abv^2$ ❌
• C: $2av^2$ ❌
• D: $-2av^3$ ✅

So the correct option is: $\boxed{D}$

5. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

They match.