Acceleration

Direction of Acceleration Describe the motion of an object with vi and a as shown to the left. Moving right as it speeds up Moving right as it slows down Moving left as it speeds up Moving left as it slows down Vi a + - To make the situations more concrete, use an automobile as an example. For example, the first combination would be a car moving to the right (v is +) and accelerating to the right (a is +), so the speed will increase. Some students may be confused by the latter two examples, thinking that a negative acceleration corresponds to slowing down and a positive acceleration corresponds to speeding up. Emphasize that the directions of velocity and acceleration must both be taken into account. In the third example, the velocity and acceleration are in the same direction, so the object is speeding up. In the fourth case, they are in opposite directions, so the object is slowing down.

Graphing Velocity The slope (rise/run) of a velocity/time graph is the acceleration. Rise is change in v Run is change in t This graph shows a constant acceleration. Average speed is the midpoint. The equation for vavg is only valid if the velocity increases uniformly (a straight line in a velocity-time graph) or, in other words, if the acceleration is constant.

Acceleration Rate of change in velocity What are the units? SI Units: (m/s)/s or m/s2 Other Units: (km/h)/s or (mi/h)/s Acceleration = 0 implies a constant velocity (or rest) Have students analyze the equation before providing the answer to the units. Stress that m/s2 are a short way of saying (m/s)/s. It is a good idea to keep saying (m/s)/s in order to emphasize the fact that acceleration is the change in velocity (m/s) over a period of time (s).

Displacement with constant acceleration ΔX = (1/2) (Vi + Vf)Δt

Displacement with Constant Acceleration Practice A car accelerates uniformly from rest to a speed of 23.7 km/h in 6.5 s. Find the displacement the car travel during this time. 21 m, forward

Velocity with constant acceleration Vf =Vi +aΔt

Displacement with constant acceleration ΔX = Vi Δt +(1/2) a (Δt)2

Displacement and Velocity with Constant Acceleration Practice A car with an initial speed of 23.7 km/h accelerates at a uniform rate of 0.92 m/s2 for 3.6 s. Find the final speed and the displacement of the car during this time. 9.9 m/s and + 30 m forward

Final Velocity after any displacement

Displacement and Velocity with Constant Acceleration Practice A car travelling initially at +7.0 m/s accelerates at the rate of +0.80 m/s2 for a distance of 245m. A. What is its velocity at the end of the acceleration? +21 m/s

Useful Equations 1. 2. 3. 4. 5. Equations (1) and (2) are the definitions of velocity and acceleration. Equations (3), (4), and (5) are only valid for uniform acceleration. Show students how to derive equation (4) by combining (1), (2), and (3). Then allow students some time to derive (5) from (1), (2), and (3) by eliminating time. Since (4) and (5) are derived from the first three, there are no problems that can be solved with them that could not have been solved by using the first three equations. It might be easier to use (4) and (5) but it is not necessary. They do not represent any “new” rules.