Rectlinear motion_JEE MAIN_12-04-2019_S2 - Competishun – Live JEE & NEET Courses and Exam Prep

JEE MAIN 2019

12-04-2019 S2

Question

A particle is moving with speed $v=b \sqrt{x}$ along positive $x$-axis. Calculate the speed of the particle at time $t=\tau$ (assume that the particle is at origin at $\mathrm{t}=0$ ).

Select the correct option:

$\mathrm{b}^2 \tau$

$\frac{b^2 \tau}{4}$

$\frac{b^2 \tau}{2}$

$\frac{b^2 \tau}{\sqrt{2}}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Physics

Medium

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