Question

An object's two dimensional velocity is given by `v(t) = ( 3t^2 - 2t , 1- 3t )`. What is the object's rate and direction of acceleration at `t=5`?

Answer 1

The acceleration is `28.2` `ms^-2` at `-84^o`: that is, at an angle in the third quadrant, downward and to the right.

Explanation:

The acceleration is the rate of change of the velocity, and is found as the first derivative of the velocity. We can treat the `x` and `y` directions independently.

`v_x=3t^2-2t`

`(dv_x)/dt=6t-2`

`v_y=1-3t`

`(dv_y)/dt=-3`

We can recombine these dimensions:

`a(t)=(6t-2,-3)`

Interestingly, the acceleration in the `y` direction is constant, while the acceleration in the `x` direction changes with time.

At `t=5` `s`, we substitute in 5, and get:

`a(t)=(6(5)-2,-3)=(28,-3)`

We can use Pythagoras' theorem to find the magnitude and the tangent to find the direction of the resultant acceleration vector.

`r=sqrt(x^2+y^2)=sqrt(28^2+(-3)^2)=sqrt(793)=28.2` `ms^-2`

`tan \theta = 28/(-3) -> \theta = -84^o`