WBJEE · Physics · Laws of Motion
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
- 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 \(\frac{a t^2}{2 m}, \frac{\mathrm{bt}^2}{2 \mathrm{~m}}, \frac{\mathrm{ct}^2}{\mathrm{~m}}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{at}^2}{2 \mathrm{~m}}, \frac{\mathrm{bt}^2}{2 \mathrm{~m}}, \frac{\mathrm{ct}^2}{2 \mathrm{~m}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \vec{F}=a \hat{i}+b \hat{j}+c \hat{k} \\ & x=\frac{1}{2}\left(\frac{a}{m}\right) t^2, y=\frac{1}{2}\left(\frac{b}{m}\right) t^2, z=\frac{1}{2}\left(\frac{c}{m}\right) t^2 \end{aligned}\)
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