if position of a particle at instant t is given by x= t3.find the acceleration of the particle..
derieve the relation for uniformly accelerated motion along the straight line. s= ut + 1/2 at2where the symbols have their usual meaing....
Consider a body having initial velocity u,If it is subjected to an acceleration 'a' so that after time 't' its velocity becomes v.
acceleration=Change in velocity/time taken
Suppose the distance travelled by the above body in time t=s
we have average velocity=(Initial velocity+Final velocity)/2
Position is given by x = t^3
We know that the second derivative of distance x gives acceleration.
So differentiate it twice, first you get velocity, then you get acceleration
First we get v = dx/dt = 3t^2
then we get acceleration , dv/dt = 6t
Make sure you know differentiation formula d(x^n) / dt = n x^(n - 1)
For the second part of your question,
We will derive the relation using average velocity
Average velocity = Total distance covered / total time taken
Let the object started with initial velocity u, after accelerating for time t, it's final velocity becomes v
then we can write acceleration a = (v - u)/t
v = u + at ----equation (1)
Also we know that the avergae velocity during this interval is the average of u and v
that is, (u + v)/2
Total distance covered = average velocity x time
s = [(u + v)/2] x t
substitute the value of v from equation 1
s = [(u + u + at)/2] x t
s = (2u + at)*t/2
s = [2ut + at^2]/2
s = ut + (1/2)at^2