Answer
For positive $t$ values.
Work Step by Step
Acceleration is defined as $$a=\frac{d^2}{dt^2}(x(t))=x''(t)$$ Taking the first derivative of position yields $v(t)=20-15t^2$. Taking the derivative of velocity yields an acceleration of $$a(t)=-30t$$ In this equation, if $t$ is positive, acceleration is negative, since a negative times a positive is a negative.
