A car travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the equation x (t) = 9t^2-8t^3 m. Find the acceleration (in m/s/s) of the car at t = 4.6 s. Give answer with two decimal places. Don't use the units in the answer box.

Given that the equation of distance "x" of a car travelling in a straight line along a road.

x (t) = 9 t² - 8 t³ meters

acceleration at time "t" = ?

t = 4.6 s

Acceleration = Velocity / time

Velocity = distance / time

Acceleration = distance / time²

Acceleration = x / t²

a (t) = (9t² - 8t³) / t²

a (t) = 9 - 8t

a (4.6) = 9 - 8 (4.6)

a (4.6) = - 27.80 m/s²

The acceleration of car at time "t" is 27.80 m/s².