## Fundamentals of Physics Extended (10th Edition)

a = -18i $m/s^2$
The velocity is given by $v = (6t-4t^2)i + 8j$ The acceleration is given by the derivative of the velocity so we need to find: $\frac{d}{dt}((6t-4t^2)i + 8j)$ We find this to be $a = (6 -4(2)t)i + 0j = (6-8t)i$ To find the acceleration at t = 3 we just need to plug in 3 for t in the expression above which gives $a = (6-8(3))i = (6-24)i = -18i$