Question

An object's two dimensional velocity is given by `v(t) = ( 1/t, t^2)`. What is the object's rate and direction of acceleration at `t=2`?

Answer 1

The acceleration at `t=2` is `a(t) = (-1/4, 4)`, whose absolute value is `sqrt257/4` and the direction is `93.576^@` w.r.t. `x`-axis.

Explanation:

Acceleration is the first derivative of velocity, so if we differentiate the velocity function, we get a function for the acceleration:

`(dv)/(dt) = (-1/t^2, 2t)`

Inserting `t=2` gives us the acceleration at `t= 2s`, namely

`a(t) = (-1/4, 4)`, whose absolute value is `sqrt((-1/4)^2+4^2)=sqrt(257/16)=sqrt257/4`

The direction is given by `alpha=tan^(-1)(4/(-1/4))=tan^(-1)(-16)=93.576^@` w.r.t. `x`-axis.