Query [from LP]: A particle is moving in a straight line and its distance (s m) from a fixed point in the line after t seconds is given as a function of t. I'm asked to find when it is instantaneously at rest?
Response: The term instantaneously is applied here because the particle only stops for a split second and then carries on moving. Simply, you are looking for when the velocity is zero.
The function of t you supplied was s(t) = 12t - 15t2 + 4t3.
The velocity function v(t), which is the first derivative of this function (i.e. s' or ds/dt) becomes... v(t) = s'(t) = 12 - 30t + 12t2 [= 0 when stationary].
Solving this quadratic gives the answer for t as 0.5 or 2 secs.
That makes sense. Ta!
Posted by: LP | August 19, 2007 at 12:37