Question

A particle's trajectory is described by x =(1/2t3?2t2)m and y =(1/2t2?2t)m, where t is in s, What is the particle's speed at t=4.5s ?

Answer 1

As we know that it is given to us

`x = ((1)/(2)t^3 - 2t^2)`

`y = ((1)/(2)t^2 - 2t)`

now we know that rate of change in position is known as velocity

so here we will have

`v_x = (dx)/(dt)`

`v_x = (3)/(2)t^2 - 4t`

similarly

`v_y = (dy)/(dt)`

`v_y = t - 2`

now we have t = 4.5 s

`v_x = 1.5(4.5)^2 - 4(4.5) = 12.375 m/s`

`v_y = 4.5 - 2 = 2.5 m/s`

now the net speed is given as

`v^2 = v_x^2 + v_y^2`

`v^2 = 12.375^2 + 2.5^2`

`v = 12.625 m/s`