Question

A toy robot is programmed to move in a straight line across a desktop according the equation x=-(11/15t^3) + 3.1t, where x is in inches and t is in seconds. At some point during the trip across the desktop, the robot is at rest. What is the robot's acceleration at this point? Select one: a. −3.5 in / s2 b. 11 in / s2 c. −5.3 in / s2 d. 13 in / s2

Answer 1

v = dx/dt = (-33/15)t^2 + 3.1

When v = 0 we need to find t => (33/15)t^2 = 3.1 => t = 1.19

a = dv/dt = (-66/15)t At t = 1.19, a = - 5.23