Question

A body moves in a plane so that the displacemenr along x and y axes given by x = 3t^3 and y =4t^3 the velocity of the body

Answer 1

Pls tell me if you find any mistakes...

I'm still in class Xth...

Answer 2

Answer:-

$\implies \: \boxed{\bf{v \: = 15 \: {t}^{2} }}$

Explanation:-

To find :-

Velocity of body.

Given:-

$\bf{x = 3{t}^{3} \: \: and \: y = 4 {t}^{3} }$

Solution :-

$\implies \: \bf{ x \: = 3 {t}^{3} }$

Differentiating with respect to" t",

$\implies \: \bf{\frac{dx}{dt} = 9 {t}^{2} } \\ \because \: \bf{v_x = \frac{dx}{dt} } \\ \therefore \: \: \bf{v_x \: = 9 {t}^{2} }$

And ,

$\implies \: \bf{y = 4 {t}^{3} }$

Differentiating with respect to "t"

$\implies \: \bf{\frac{dy}{dt} = 12 {t}^{2} } \: \\ \because \: \bf{ v_y = \frac{dy}{dx} } \: \\ \therefore \: \bf{v_y \: = 12 {t}^{2} }$

Hence,

$\implies \: \bf{ v = \sqrt{ { \: v_x \: }^{2} + { v_y}^{2} } } \\ \\ \implies \: \bf{v = \sqrt{81 \: {t}^{4} +144 \: {t}^{4} } } \\ \\ \implies \: \bf{v = \sqrt{225 \: {t}^{4} } } \\ \\ \implies \: \boxed{ \bf{ v = 15 \: {t}^{2} }}$

Hope it helps you.