Q133 / 160 · WBJEE 2009 · Physics

The acceleration ' $a$ ' (in $\mathrm{ms}^{-2}$ ) of a body, starting from rest varies with time $t$ (in $s$ ) following the equation $a=3 t+4$ The velocity of the body at time $t=2 \mathrm{~s}$ will be

  • A
    $10 \mathrm{~ms}^{-1}$
  • B
    $18 \mathrm{~ms}^{-1}$
  • Correct answer
    C
    $14 \mathrm{~ms}^{-1}$
  • D
    $26 \mathrm{~ms}^{-1}$

Solution

$a=3 t+4$ $\begin{aligned} & \frac{d V}{d t}=3 t+4 \\ & \int_0^V d V=\int_0^t(3 t+4) d t \\ & V=\frac{3 t^2}{2}+4 t=\frac{12}{2}+8=14 \mathrm{~m} / \mathrm{s}\end{aligned}$

Hence, the correct answer is option: 'C'.

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