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. This, instantaneous acceleration is

    instantaneous acceleration in two dimensions
  • 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.

Concept Map's for Velocity and acceleration in two dimensional motion

Concept Map for Velocity and acceleration in two dimensional motion
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
It is given in the questions
r= 3ti + 2t2j+ 11k
v(t) = dr/dt
v(t) =d[3ti + 2t2j+ 11k ]/dt =3i +4tj
a(t) = dv/dt
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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