Acceleration: acceleration is the rate of change of velocity.

• Acceleration is zero if an object moving at constant velocity.
  
• Acceleration is positive if an object is moving at increasing velocity.
  
• Acceleration is negative if an object is moving at decreasing velocity.

Average Acceleration: total change in velocity DV during a time interval Dt,

Average Acceleration = <a> = DV /Dt = (V2 - V1)/(T2 - T1) = (Vf - Vi)/(Tf - Ti)

Figure 1

If the magnitude of V1 is V1 = 2 m/s at t=T1 and the magnitude of V2 is V2 = 3 m/s at t=T2 =T1 + 5s, determine the average acceleration (magnitude and direction)

Figure 1 shows how DV is determined graphically: its direction is as shown and its magnitude can be measured using a ruler. We can calculate its magnitude from the Law of Cosines: (DV)**2 = V1**2 + V2**2 - 2V1 V2 cos a, where a is the angle between the vectors V1 and V2. The angle a can be measured from the diagram, giving a ª 63 degrees.

The result is DV = (4 + 9 - 12 cos 63)**0.5 = 2.75 m/s.

The magnitude of the average acceleration is (2.75 m/s) / 5s = 0.55 m/s**2. The direction of the average acceleration is the same as the direction of DV shown in Figure 1.

Instantaneous Acceleration: acceleration at a particular instant of time, i.e.

    a = (DV/Dt) as Dt approaches zero = dV/dt
       = the slope of the velocity-time curve at any instant.

Uniformly Accelerated Motion: Constant Acceleration

Figure 2 shows the trajectory of a baseball with an initial velocity Vi in a cartesian coordinate system where g is the gravitationl acceleration. Calculate the velocity of the baseball as a function of time.
  

Exanmple: Acceleration by gravity - free fall neglecting the air drag a =(Vf - Vi)/(Tf - Ti)    = const. (Vf - Vi) = a(Tf - Ti)              = a t     Vf = Vi + a t  .........(1)

Figure 2

Given Vi = (10, 0, 12) m/s, the velocity V(t) = (Vx, Vy, Vz) can be calculated as

Vx(t) = 10 m/s,     Vy(t) = 0 m/s,     Vz(t) = (12 - 9.8 t) m/s

What is the range of the ball (when the ball hit the ground, i.e., z = 0)?

    z = 12 t - (1/2) 9.8 t**2 = 0, solve for t.   

    One obtains t = 0 and t = 24/9.8 = 2.45 s

    The range is x = 10 t = 10x2.45 = 24.5 m

What is the maximum height reached by the baseball?

     H(max) is reached when Vz = 12 - 9.8 t = 0, i.e., t = 12/9.8 = 1.22 s

     so that H(max) = z(t=1.22) = [12 - (1/2) 9.8 t] t = 7.35 m

What is the instantaneous velocity of the ball at the maximum height?

     V(at max height) = (10, 0, 0) m/s

Quiz # 2 turned into a lecture of Chapter 3:

A kangaroo jumps straight up to reach a height of 2.5 m. The z axis is taken positive upward.

(a) From the definitions for the acceleration and the velocity, derive the speed v and position      z of the motion under the influence of the gravity:     
             V = Vi - gt    and     z = zi + Vi t - (1/2) gt**2
       where vi is the initial velocity, the minus sign in the equation comes from gravitational      acceleration g being downward with the magnitude of 9.8 m/s**2.

(b) Use the results in (a) to calculate Kangaroo's take-off velocity.

(c) Use (a) to calculate its landing velocity if the kangaroo starts from 2.5 m height at zero velocity.
----------------------------------- Solution ------------------------------------
(a)  aav = (V - Vi)/(t-ti) ---- definition for acceleration. Set ti = 0, one obtains
       V = Vi + aav t, and aav = (1/2)(af + ai) = (1/2)[(-g) + (-g)] = -g              

V = Vi - gt ------------(1)

       Vav = (z - zi)/(t-ti) ----- definition of velocity along the z-axis.
       z = zi + Vav t = zi + (1/2)(V + Vi) t = zi + (1/2)Vt + (1/2)Vit
          = zi + (1/2)(Vi - gt)t + (1/2)Vit

z = zi + Vit - (1/2)gt**2 ------(2)

(b)  V = 0,  zi = 0, z = 2.5 m,     Vi = ?    t = ?
       Using the conditions given above, equations (1) and (2) can be reduced to
                      Vi = gt ------------------------- (1') 
                       2.5 = Vit - (1/2)gt**2 ---------- (2')
        Substituting (1') for Vi into (2'), one obtains 2.5 = (1/2)gt**2 so that
        t = (5/g)**0.5 = 0.714 s   and    Vi = gt = 7.0 m/s

(c)  zi = 2.5 m,   z = 0,     Vi = 0    V = ?
         The above conditions reduces equations (1) and (2) to
                        V = 0 - gt -------------(1'')
                        0 = 2.5 - (1/2)gt**2 ------(2'')
       Solving for t from (2") and putting it into (1"), one obtains
        t = (2.5 x2/g)**0.5 = 0.714 s    and     V = -7.0 m/s.