Question

The coordinates of a moving particle at any time t, are given by $x=a{{t}^{2}}$and $y=b{{t}^{2}}$. The speed of the particle is?
(A) $2t(a+b)$
(B) $2t\sqrt{({{a}^{2}}+{{b}^{2}})}$
(C) $2t\sqrt{({{a}^{2}}-{{b}^{2}})}$
(D) $\sqrt{({{a}^{2}}+{{b}^{2}})}$

Answer 1

Hint:We know that speed is a scalar quantity. This means that it has only magnitude and no direction. Speed can be defined as the rate of change of distance. While calculating the speed we take the ratio of the total distance covered to the total time taken. But here we are given the coordinates and we need to find the speed. We can use differential calculus to find the speed by differentiating the two coordinates with respect to the time.

Complete step by step answer:
The x coordinate is $x=a{{t}^{2}}$, so find the x coordinate of the speed,
$\Rightarrow \dfrac{dx}{dt}=\dfrac{d(a{{t}^{2}})}{dt}$
$\Rightarrow \dfrac{dx}{dt}=2at$
$\Rightarrow {{v}_{x}}=2at$--(1)
Now for y coordinate,
$\Rightarrow \dfrac{dy}{dt}=\dfrac{d(b{{t}^{2}})}{dt}$
$\Rightarrow \dfrac{dy}{dt}=2bt$
$\Rightarrow {{v}_{y}}=2bt$--(2)
Since, the x coordinate and the y coordinate are perpendicular to each other, thus,
$\Rightarrow v=\sqrt{v_{x}^{2}+v_{y}^{2}}$
$\Rightarrow v=\sqrt{{{(2at)}^{2}}+{{(2bt)}^{2}}}$
$\Rightarrow v=\sqrt{4{{t}^{2}}({{a}^{2}}+{{b}^{2}})}$
$\therefore v=2t\sqrt{({{a}^{2}}+{{b}^{2}})}$

So, the correct option is B.

Note:Most of the time we forget to use the coordinates system. But in physics they play a very important role. Speed and velocity both are sometimes taken in the same context but both are entirely different quantities. Velocity us a vector quantity and it has both the magnitude and the direction. The magnitude refers to the amount or the quantity of any body. The use of differential calculus plays a very significant role in physics. When we have to take a ratio of two quantities but in the question only the variables are mentioned then it comes into play.