The position vector of a moving body at any instant of time is given as vec r = (5t^2 i - 5tj)m.

Question

The position vector of a moving body at any instant of time is given as \(\vec{\mathrm{r}}=\left(5 \mathrm{t}^{2} \hat{\mathrm{i}}-5 \mathrm{t} \hat{\mathrm{j}}\right) \mathrm{m}\). The magnitude and direction of velocity at \(\mathrm{t}=2 \mathrm{s}\) is,

(1) \(5 \sqrt{15} \mathrm{~m} / \mathrm{s}\), making an angle of \(\tan ^{-1} 4\) with -ve Y axis

(2) \(5 \sqrt{15} \mathrm{~m} / \mathrm{s}\), making an angle of \(\tan ^{-1} 4\) with +ve X axis

(3) \(5 \sqrt{17} \mathrm{~m} / \mathrm{s}\), making an angle of \(\tan ^{-1} 4\) with -ve Y axis

(4) \(5 \sqrt{17} \mathrm{~m} / \mathrm{s}\), making an angle of \(\tan ^{-1} 4\) with +ve X axis

Answer 1

Correct option is : (3) \(5 \sqrt{17} \mathrm{~m} / \mathrm{s}\), making an angle of \(\tan ^{-1} 4\) with -ve Y axis

\(\vec{\mathrm{r}}=5 \mathrm{t}^{2} \hat{\mathrm{i}}-5 \hat{\mathrm{j}}\) 

\( \begin{aligned} & \vec{v}=10 t \hat{i}-5 \hat{j} \end{aligned} \) 

\( \begin{aligned} \vec{v}=20 \hat{i}-5 \hat{j} \end{aligned} \)  

at t = 2sec

 

\( \tan \theta=\frac{20}{5}=4\) 

\( \theta=\tan ^{-1} 4\)

From –ve Y-axis