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.
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