Question

The coordinates of a moving particle at time t are given by x=c t^2 and y=b t^2. The speed of particle is given by

Answer 1

Ans. is v = 2t ✓(a^2+b^2) .... solution is in the pic :)

Answer 2

Answer:

$v=2t\sqrt{c^2+b^2}$

Explanation:

The x- coordinate of a moving particle is given by :

$x=ct^2$

Velocity in x direction is given by :

$v_x=\dfrac{dx}{dt}=2ct$................(1)

The y- coordinate of a moving particle is given by :

$y=bt^2$

Velocity in x direction is given by :

$v_y=\dfrac{dy}{dt}=2bt$..............(2)

Let v is the speed of the particle which is given by :

$v=\sqrt{v_x^2+v_y^2}$

$v=\sqrt{(2ct)^2+(2bt)^2}$

$v=\sqrt{4t^2(c^2+b^2)}$

$v=2t\sqrt{c^2+b^2}$

So, the speed of particle is $2t\sqrt{c^2+b^2}$. Hence, this is the required solution.