4. Average and instantaneous acceleration

4. Average and instantaneous acceleration

• Suppose a particle moves from point P to point Q in x-y plane as shown below in the figure
  
  
  
  
• Suppose v₁ is the velocity of the particle at point P and v₂ is the velocity of particle at point Q
• Average acceleration is the change in velocity of particle from v₁ to v₂ in time interval Δt as particle moves from point P to Q. Thus average acceleration is
  
   
  Average accelaration is the vector quantity having direction same as that of Δv.
• Again if point Q aproaches point P, then limiting value of average acceleration as time aproaches zero defines instantaneous acceleration or simply the acceleration of particle at that point. Ths, instantaneous acceleration is 
• Figure below shows instantaneous acceleration a at point P.
  
  
   
  
• Instantaneous acceleration does not have same direction as that of velocity vector instead it must lie on the concave side of the curved surface.
• Thus velocity and acceleration vectors may have any angle between 0 to 180 degree between them.

Question
The position of a object is given by
r= 3ti + 2t²j+ 11k 
Where t is in second and coefficents have the proper units for r to be in centimeters
a) Find v(t) and a(t) of the object
b) Find the magnitude and direction of the velocity at t=3sec
Solution
It is given in the questions
r= 3ti + 2t²j+ 11k 
Now
v(t) = dr/dt
Therefore,
v(t) =d[3ti + 2t²j+ 11k ]/dt =3i +4tj
Now
a(t) = dv/dt
Therefore,
a(t) =d[3i +4tj]/dt =4j
So acceleration is 4 cm/s² across y-axis

Now velocity at 3 sec

v(t) =3i +4tj=3i +12j
So its magnitude is √(3² +12²) = 12.4 cm/s
And direction will be tan^-1(v^y/v^x) =tan^-1(4) =76⁰