Q. The position vector of a moving body at any instant of time is given as $\vec r=\left(5t^2\hat i-5t\hat j\right)$ m. The magnitude and direction of velocity at $t=2$ s is,
A) $5\sqrt{15}$ m/s, making an angle of $\tan^{-1}4$ with -ve $Y$ axis
B) $5\sqrt{15}$ m/s, making an angle of $\tan^{-1}4$ with +ve $X$ axis
C) $5\sqrt{17}$ m/s, making an angle of $\tan^{-1}4$ with +ve $X$ axis
D) $5\sqrt{17}$ m/s, making an angle of $\tan^{-1}4$ with -ve $Y$ axis
Explanation
[Solution by Dr. Abubacker] $\vec v=\frac{d\vec r}{dt}=(10t\,\hat i-5\hat j)$. At $t=2$ s: $\vec v=(20,-5)$, $|\vec v|=\sqrt{425}=5\sqrt{17}$ m/s. The angle with the $-Y$ axis satisfies $\tan\theta=\frac{20}5=4$, i.e. $\tan^{-1}4$. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.