Question

The motion of the particle is given parametrically by `x(t)=3t^2-3`, `y(t)=2t^2` for t is greater than or equal to 0, how do you find the speed of the particle at t=1?

Answer 1

The speed is constant for all `t` and is `4/6` units per second

Explanation:

`dx/dt = 6t`

`dy/dt = 4t`

The speed is `dy/dx`

`dy/dx = (dy/dt)/(dx/dt) = (4t)/(6t) = 4/6`