Question

The position equation for a particle is where s is measured in feet and t is measured in seconds. Find the acceleration of the particle at 4 seconds?

`s(t)=2t^2+sqrt(t)`

Answer 1

`S=2t^2 +sqrt(t)`

So, Velocity=`(ds)/(dt)=4t + 1/2 t^(-1/2)`

And,acceleration= `(dv)/(dt)=4 -1/4 t ^(-3/2)`

So, acceleration,at `t=4` is, `4- 1/4 (2^2)^(-3/2) = 4 -(1/4)(1/8)=3.96 fts^-2`