Question

The coordinates of moving particles at time t are given by $x=ct^2$and $y=bt^2$. The speed of the particle is given by:
$\text{A.}\quad 2t(c+b)$
$\text{B.}\quad 2t\sqrt{c^2-b^2}$
$\text{C.}\quad t\sqrt{c^2+b^2}$
$\text{D.}\quad 2t\sqrt{c^2+b^2}$

Answer 1

There are majorly two types of quantities, scalar and vector quantities. All the quantities are divided into these two categories. Scalar quantities are those quantities, which has only magnitude e.g. – mass, speed, pressure, etc. Vector quantities are those which has both magnitude and directions eg – weight, velocity and thrust. For instance, distance is a scalar quantity whereas displacement is a vector quantity. Velocity could be found by differentiating the displacement with respect to time and speed is the magnitude of velocity.
Formula used:
$v_x = \dfrac{dx}{dt},\ v_y = \dfrac{dy}{dt}$

Complete answer:
How do we measure velocity?
Velocity is a vector quantity. To understand the velocity, we have to look for both magnitude and resultant of the velocity vector. It is also defined as the rate of change of displacement.
$v = \dfrac{ds}{dt}$, where ‘s’ is the displacement.
How do we measure speed?
Speed is a scalar quantity. It is also defined as the rate of change of distance.
It is measured by taking the magnitude of velocity.
Now, $v_x = \dfrac{dx}{dt},\ v_y = \dfrac{dy}{dt}$
$v_x = \dfrac{d}{dt}(ct^2) = 2ct$
$v_y = \dfrac{d}{dt} (bt^2) = 2bt$
Now, $R = \sqrt{A^2+B^2+2ABcos\theta}$, where R is the magnitude of two vectors A and B.
Since angle between x and y-axis is $90^{\circ}$
So $R = \sqrt{A^2+B^2}$
Or $v_{net} = \sqrt{v_x^2+v_y^2}$
$v_{net} = \sqrt{(2ct)^2 + (2bt)^2} = 2t\sqrt{b^2+c^2}$
Now, this is the magnitude of velocity.
Which is also equal to the speed also. Thus speed is $2t\sqrt{b^2+c^2}$.

So, the correct answer is “Option D”.

Note:
In the above expression, we have the magnitude $v_{net} = \sqrt{v_x^2+v_y^2}$, which has nothing to do with the direction of$v_{net}$i.e. resultant whereas in expression,$\alpha =tan^{-1}\left( \dfrac{v_y }{v_x}\right)$ we have nothing to do with the magnitude of$\vec R$. If we want to represent the velocity vector, we also have to provide the direction of net velocity also along with the magnitude. $\alpha =tan^{-1}\left( \dfrac{v_y }{v_x}\right) tan^{-1}\left( \dfrac{2bt }{2ct}\right) = tan^{-1} \left( \dfrac bc \right)$. Hence velocity has magnitude $2t\sqrt{b^2+c^2}$and is along $\alpha = tan^{-1} \dfrac bc$with the horizontal.