Q154 / 225 · JEE Main 2005 · Physics
The relation between time $t$ and distance $\mathrm{x}$ is $\mathrm{t}=a \mathrm{x}^2+\mathrm{bx}$ where $\mathrm{a}$ and $\mathrm{b}$ are constants. The acceleration is
- A$-2 a b v^2$
- B$2 b v^3$
- Correct answerC$-2 a v^3$
- D$2 a v^2$
Solution
$$
t=a x^2+b x
$$
by differentiating acceleration $=-2 a v^3$
Hence, the correct answer is option: 'C'.