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

`2 sqrt(10)`

Explanation:

`x = 3t^2 - 3, dot x = 6t`
`y = 2t _ 2, dot y = 2`

where `dot x, dot y` are the velocities in the x and y directions

speed s is the magnitude of the velocity vector and so `s= sqrt (dot x^2 + dot y ^2 )`

at `t = 1, dot x = 6, dot y = 2` so

`s = sqrt(40) = 2 sqrt(10)`