Q

# Help me answer: - Kinematics - JEE Main-2

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

• Option 1)

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

• Option 2)

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

• Option 3)

$b^{2}\tau$

• Option 4)

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

Views

$V=b\sqrt{x}$

$\frac{dx}{dt}=b\sqrt{x}$

$\int _{o}^{x}\frac{dx}{\sqrt{x}}=b\int ^{t}_{o}dt$

$2\sqrt{x}=bt$

$x=\frac{b^{2}t^{2}}{4}\cdots (a)$

$\frac{dx}{dt}=\frac{b^2t}{2}\cdots (b)$

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

Option 1)

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

Option 2)

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

Option 3)

$b^{2}\tau$

Option 4)

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

Exams
Articles
Questions