Q132 / 155 · WBJEE 2025 · Physics

A force $\overrightarrow{\mathrm{F}}=a \hat{i}+b \hat{\mathrm{j}}+c \hat{k}$ is acting on a body of mass $m$. The body was initially at rest at the origin. The co-ordinates of the body after time ' t ' will be

  • Correct answer
    A
    $\frac{\mathrm{at}^2}{2 \mathrm{~m}}, \frac{\mathrm{bt}^2}{2 \mathrm{~m}}, \frac{\mathrm{ct}^2}{2 \mathrm{~m}}$
  • B
    $\frac{\mathrm{at}^2}{2 \mathrm{~m}}, \frac{\mathrm{bt}{ }^2}{\mathrm{~m}}, \frac{\mathrm{ct}^2}{2 \mathrm{~m}}$
  • C
    $\frac{a t^2}{m}, \frac{\mathrm{bt}^2}{2 \mathrm{~m}}, \frac{\mathrm{ct}^2}{2 \mathrm{~m}}$
  • D
    $\frac{a t^2}{2 m}, \frac{\mathrm{bt}^2}{2 \mathrm{~m}}, \frac{\mathrm{ct}^2}{\mathrm{~m}}$

Solution

$\begin{aligned} & \vec{F}=a \hat{i}+b \hat{j}+c \hat{k} \\ & x=\frac{1}{2}\left(\frac{a}{m}\right) t^2, y=\frac{1}{2}\left(\frac{b}{m}\right) t^2, z=\frac{1}{2}\left(\frac{c}{m}\right) t^2 \end{aligned}$

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

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