# Determine the speed of an object moving along the path described by x=3-2t^2 y=t^2 when t=1/2

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First lets consider the x direction.

`x = 3 - 2t^2`

velocity in the x direction is given by `dx/dt`

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hence lets derive it

`dx/dt = 0-4t = -4t`

` `when t = 1/2, `dx/dt = -2`

therefore the velocity in the x direction is -2 when t = 1/2

lets consider the y direction

`y = t^2`

velocity in the y direction is given by `dy/dt`

so lets derive it,

`dy/dt = 2t`

when t = 1/2, `dy/dt = 1`

therefore the velocity in y direction is 1 when t = 1/2

lets consider speed as s,

`s = sqrt((dx/dt)^2+(dy/dt)^2))`

`s = sqrt((-2)^2 + 1^2)`

`s = sqrt(5)`

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` `therefore the speed is `sqrt(5)`

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You need to remember that the speed is a scalar quantity and velocity is a vector quantitym, hence you should evaluate the velocity vector first such that:

`v(t) = (x'(t) , y'(t))`

`v(t) = (-4t , 2t)`

You need to evaluate the velocity at `t = 1/2` .

`v(1/2) = (-4/2 , 2/2) =gt v(1/2) = (-2,1)`

You should remember that the speed is the magnitude of velocity vector such that:

`s = sqrt((-2)^2 + 1^2)`

`s = sqrt(4+1) =gt s = sqrt5`

Hence, evaluating the speed of the object, moving along the given path, at t = 1/2, yields `s(1/2) = sqrt 5` .