Question

The coordinates of a particle moving in $x-y$ plane at any instant of time $t$ are $x = 4t^2 ; y = 3t^2$. The speed of the particle at that instant is

• A. 10 t
• B. 5 t
• C. 3 t
• D. 2 t
• E. $\sqrt{13}\, t $

Answer 1

Answer: (A)

$x=4 t^{2}$
Differentiate with respect to $t$, we get
$v_{x}=\frac{d x}{d t}=2 \times 4 \,t=8 \,t $
Similarly, $ v_{Y}=\frac{d y}{d t}=2 \times 3\, t=6\, t$
The speed of the particle
$V =\sqrt{V_{X}^{2}+V_{Y}^{2}} $
$=\sqrt{(8\, t)^{2}+(\overline{6} t)^{2}} $
$=\sqrt{61\, t^{2}+36\, t^{2}} $
$=\sqrt{100\, t^{2}}=1 0\,t $