The velocity of a particle moving along the x-axis varies according to the expression vx=(60 -0.72) m/s, where vy is in meters per second and t is in seconds. (m/s) 0 1 2 3 4 t(s) The velocity-time graph for a particle moving along the x-axis. Think about what the particle is doing from the mathematical representation. Is it moving at t=0? Yes No In which direction? in the positive direction in the negative direction the particle is not moving As the particle moves from A to B, does it speed up or slow down? It speeds up. It slows down. The figure is a vyt graph that was created from the velocity versus time expression given in the problem statement. Because the slope of the entire vx - t curve is negative, we expect the acceleration to be --Select-- (a) Find the average acceleration in the time interval t = 0 to t = 2.0 s. SOLUTION Find the velocities (in m/s) at t, t, and ty ta 2.0 s by substituting these values of t into the expression for the velocity: VXA= m/s m/s Find the average acceleration (in m/s²) in the specified time interval Atta - 2.0 s (Indicate the direction with the sign of your answer.): 3x, avg t-t to-[A m/s² The negative sign is consistent with our expectations-namely, that the average acceleration, represented by the slope of the line joining the initial and final points on the ---Select-- graph, is negative. (b) Determine the acceleration at t = 2.0 s. SOLUTION (Use the following as necessary: t and At.). Knowing that the initial velocity at any time t is vx (60 -0.72) m/s, find the velocity at any later time t + At: m/s Find the change in velocity over the time interval At: m/s To find the acceleration at any time t, divide this expression by At and take the limit of the result as At approaches zero: alim Δεν Δε το Δε m/s² Substitute t = 2.0 s (Enter your answer in m/s². Indicate the direction with the sign of your answer.): m/s² Because the velocity of the particle is --Select--and the acceleration is -Select- at this instant, the particle is --Select-- Notice that the answers to parts (a) and (b) are different. The average acceleration in (a) is the slope of the blue line in the figure above connecting points A and B. The instantaneous acceleration in (b) is the slope of the green line tangent to the curve at point B. Notice also that the acceleration is not constant in this example. Situations involving constant acceleration are treated later.