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 answerA$\frac{\mathrm{at}^2}{2 \mathrm{~m}}, \frac{\mathrm{bt}^2}{2 \mathrm{~m}}, \frac{\mathrm{ct}^2}{2 \mathrm{~m}}$
- 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'.