JEE Mains · Physics · STD 11 - Motion in plane
- JEE Mains 2025
- Medium
- MCQ
The position vector of a moving body at any instant of time is given as \(\vec{r}=\left(5 t^2 \hat{i}-5 t \hat{j}\right) \mathrm{m}\). The magnitude and direction of velocity at \(t=2 \mathrm{~s}\) is,
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Answer & Solution
Correct Answer
(D) \(5 \sqrt{17} \mathrm{~m} / \mathrm{s}\), making an angle of \(\tan ^{-1} 4\) with - ve \(Y\) axis
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \overrightarrow{\mathrm{r}}=5 t^2 \hat{\mathrm{i}}-5 \mathrm{t}_{\mathrm{j}} \\ & \overrightarrow{\mathrm{v}}=10 \hat{\mathrm{i}}-5 \hat{\mathrm{j}} \\ & \overrightarrow{\mathrm{v}}=20 \hat{\mathrm{i}}-5 \hat{\mathrm{j}} \quad \text { at } \mathrm{t}=2 \mathrm{sec}\end{aligned}\)
\(\begin{aligned} & \tan \theta=\frac{20}{5}=4 \\ & \theta=\tan ^{-1} 4 \\ & \text { From -veY-axis }\end{aligned}\)
\(\begin{aligned} & \tan \theta=\frac{20}{5}=4 \\ & \theta=\tan ^{-1} 4 \\ & \text { From -veY-axis }\end{aligned}\)
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