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Average and instantaneous acceleration in two dimensions




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


    Average accleration vector representation in two dimensions

  • 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 accleration in two dimensions

    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 at a point

  • 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(vy/vx) =tan-1(4) =760

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