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 answer
    C
    $-2 a v^3$
  • 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'.

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