During the time interval 0 < t < 1, the velocity of an object as a function of time is v = 5 + 3t^2 + 2t^3. Knowing that the object follows a rectilinear trajectory, What is the correct expression for the acceleration?

A dog running in an open field has components of velocity vx = 2.6 m/s and vy = -1.2 m/s at time t1 = 11.7 s . For the time interval from t1 = 11.7 s to t2 = 22.8 s, the average acceleration of the dog has magnitude 0.49 m/s^2 and direction 25.0  measured from the +x−axis toward the +y−axis. At t2, what is the x component, y component, and magnitude of the dog's velocity? What is the direction of the dogs velocity?

A particle moves along the positive x-axis and its position as a function of time is x(t)=(19.9 m/s)t − (1.7 m/s^2)t^2. What is the average velocity in the time interval between 2.3s and 3.7s in m/s?

Suppose that the position of a particle as a function of time is given by the expression: x(t) = (1t4 + 3t²) Î + 2t5ĵ Determine the velocity as a function of time, v(t) = Determine the acceleration as a function of time, a(t) = Determine the direction of the velocity at t = 1.4, 0v(t=1.4) = î + î+ degrees Ĵ Ĵ