# The position of a body moving along a coordinate line is given by the function s=sint+cost where s is in meters and t is in seconds . Find the following at time t= pi/3 seconds. a) find the velocity b) the speed c) the acceleration

## Expert Answers

You need to differentiate the position function with respect to t to find the velocity of particle such that:

`(ds)/(dt) = cos t - sin t`

You need to find velocity at `t = pi/3` , hence, you should substitute `pi/3 ` for t in equation `(ds)/(dt) = cos t - sin t`  such that:

`(ds)/(dt)|_(t=pi/3) = cos pi/3- sin pi/3`

`(ds)/(dt)|_(t=pi/3) = 1/2 - sqrt3/2`

`(ds)/(dt)|_(t=pi/3) = (1-sqrt3)/2`

Hence, evaluating the velocity at `t = pi/3`  yields `(ds)/(dt)|_(t=pi/3) = (1-sqrt3)/2` .

b) You need to evaluate the speed at `t = pi/3` , such that:

speed = `|(1-sqrt3)/2| = (sqrt3-1)/2`

Hence, evaluating the speed at `t = pi/3`  yields speed `= |(1-sqrt3)/2| = (sqrt3-1)/2` .

c) You need to differentiate the velocity function with respect to t to find the acceleration of particle such that: ` (dv)/(dt) = -sin t - cos t`

You need to find acceleration at `t = pi/3` , hence, you should substitute `pi/3 ` for t in equation `(dv)/(dt) =-sin t - cos t`  such that:

`(dv)/(dt)|_(t=pi/3) = -sin pi/3 - cos pi/3`

`(dv)/(dt)|_(t=pi/3) = -sqrt3/2 - 1/2`

`(dv)/(dt)|_(t=pi/3) = -(sqrt 3 + 1)/2`

Hence, evaluating the acceleration at `t = pi/3`  yields `(dv)/(dt)|_(t=pi/3) = -(sqrt 3 + 1)/2.`

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