Question

The x and y displacement of a particle in the x-y plane at any instant are given by x = ${aT}^2$ and y = 2aT where a is a constant. The velocity of the particle at any instant is given by

• A. $4a\sqrt{T^2+4}$
• B. $2a\sqrt{T^2+1}$
• C. $4a\sqrt{T^2+1}$
• D. $\frac{a}{2}\sqrt{T^2+4}$

Answer 1

The correct option is B $2a\sqrt{T^2+1}$

Given: $x=a{T}^2$ ; $y=2aT$

Differentiate both sides with respect to T;

$x=a{T}^2$

$\frac{dx}{dT}=a\times \frac{d}{dT}\left(T^2\right)$

$v_x=\frac{dx}{dT}=2aT......Eq:01$

Differentiate both sides with respect to T;

$y=2aT$

$\frac{dy}{dT}=2a\times \frac{d}{dT}\left(T\right)$

$v_y=\frac{dy}{dT}=2a......Eq:02$

Resultant velocity (V):

$V^2=(v_x{)}^2+(v_y{)}^2$

$V^2=(2aT{)}^2+(2a{)}^2$

$V^2=4a^2{T}^2+4a^2$

$V^2=4a^2(T^2+1)$

$V=2a\sqrt{T^2+1}$

Option (B) is correct.